D(cot 2)= (-csc2). It is often useful to create a visual representation of Equation for the chain rule. Jump down to problems and their solutions. The outer function in this example is “tan.” (Note: Leave the inner function in the equation (√x) but ignore that too for the moment) The derivative of tan x is sec2x, so: However, the reality is the definition is sometimes long and cumbersome to work through (not to mention it’s easy to make errors). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] The chain rule can be used to differentiate many functions that have a number raised to a power. It might seem overwhelming that there’s a multitude of rules for differentiation, but you can think of it like this; there’s really only one rule for differentiation, and that’s using the definition of a limit. That’s what we’re aiming for. Get complete access: LOTS of problems with complete, clear solutions; tips & tools; bookmark problems for later review; + MORE! Then you would next calculate $10^7,$ and so $(\boxed{\phantom{\cdots}})^7$ is the outer function. The second is more formal. We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = x^7 – 4x^3 + x.$ Then $f'(u) = e^u,$ and $g'(x) = 7x^6 -12x^2 +1.$ Hence \begin{align*} f'(x) &= e^u \cdot \left(7x^6 -12x^2 +1 \right)\\[8px] = 2(3x + 1) (3). &= \dfrac{1}{2}\dfrac{1}{ \sqrt{x^2+1}} \cdot 2x \quad \cmark \end{align*}, Solution 2 (more formal). Tip You can also use this rule to differentiate natural and common base 10 logarithms (D(ln x) = (1/x) and D(log x) = (1/x) log e. Multiplied constants add another layer of complexity to differentiating with the chain rule. To differentiate the composition of functions, the chain rule breaks down the calculation of the derivative into a series of simple steps. Differentiating using the chain rule usually involves a little intuition. \left[\left(x^2 + 1 \right)^7 (3x – 7)^4 \right]’ &= \left[ \left(x^2 + 1 \right)^7\right]’ (3x – 7)^4\, + \,\left(x^2 + 1 \right)^7 \left[(3x – 7)^4 \right]’ \\[8px] : ), this was really easy to understand good job, Thanks for letting us know. Let’s use the second form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} }\] equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). : ), What a great site. The chain rule in calculus is one way to simplify differentiation. Just ignore it, for now. Even though we had to evaluate f′ at g(x)=−2x+5, that didn't make a difference since f′=6 not matter what its input is. The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. In this case, the outer function is the sine function. What’s needed is a simpler, more intuitive approach! By continuing, you agree to their use. With some experience, you won’t introduce a new variable like $u = \cdots$ as we did above. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Covered for all Bank Exams, Competitive Exams, Interviews and Entrance tests. For example, to differentiate Solution: d d x sin( x 2 os( x 2) d d x x 2 =2 x cos( x 2). Chain Rule Example #1 Differentiate $f(x) = (x^2 + 1)^7$. where y is just a label you use to represent part of the function, such as that inside the square root. There are lots more completely solved example problems below! Step 1 Differentiate the outer function first. Example: Differentiate y = (2x + 1) 5 (x 3 – x +1) 4. Watch the video for a couple of chain rule examples, or read on below: The formal definition of the chain rule: Solution. It is useful when finding the … rule d y d x = d y d u d u d x ecomes Rule) d d x f ( g ( x = f 0 ( g ( x )) g 0 ( x ) \outer" function times of function. It’s more traditional to rewrite it as: We have the outer function $f(u) = \tan u$ and the inner function $u = g(x) = e^x.$ Then $f'(u) = \sec^2 u,$ and $g'(x) = e^x.$ Hence \begin{align*} f'(x) &= \sec^2 u \cdot e^x \\[8px] Step 1 Differentiate the outer function. dF/dx = dF/dy * dy/dx Example: Find d d x sin( x 2). Click HERE for a real-world example of the chain rule. Step 4 Great problems for practicing these rules. Label the function inside the square root as y, i.e., y = x2+1. That isn’t much help, unless you’re already very familiar with it. Use the chain rule to differentiate composite functions like sin(2x+1) or [cos(x)]³. The derivative of 2x is 2x ln 2, so: Identify the mistake(s) in the equation. 7 (sec2√x) / 2√x. Remember that a function raised to an exponent of -1 is equivalent to 1 over the function, and that an exponent of ½ is the same as a square root function. Consider a composite function whose outer function is $f(x)$ and whose inner function is $g(x).$ The composite function is thus $f(g(x)).$ Its derivative is given by: \[\bbox[yellow,8px]{ \begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\], Alternatively, if we write $y = f(u)$ and $u = g(x),$ then \[\bbox[yellow,8px]{\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} }\]. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Solution 2 (more formal). You can find the derivative of this function using the power rule: Find the derivatives equation: az dx2 : az дхду if z(x,y) is given by the ay 23 + 3xyz = 1 3. In this example, the inner function is 3x + 1. Use the chain rule to calculate h′(x), where h(x)=f(g(x)). Hint : Recall that with Chain Rule problems you need to identify the “ inside ” and “ outside ” functions and then apply the chain rule. Hyperbolic Functions - The Basics. The Chain Rule is a big topic, so we have a separate page on problems that require the Chain Rule. We have the outer function $f(u) = u^8$ and the inner function $u = g(x) = 3x^2 – 4x + 5.$ Then $f'(u) = 8u^7,$ and $g'(x) = 6x -4.$ Hence \begin{align*} f'(x) &= 8u^7 \cdot (6x – 4) \\[8px] &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] Show Solution. = e5x2 + 7x – 13(10x + 7), Step 4 Rewrite the equation and simplify, if possible. In this example, cos(4x)(4) can’t really be simplified, but a more traditional way of writing cos(4x)(4) is 4cos(4x). &= \sec^2(e^x) \cdot e^x \quad \cmark \end{align*}, Now let’s use the Product Rule: \[ \begin{align*} (f g)’ &= \qquad f’ g\qquad\qquad +\qquad\qquad fg’ \\[8px] Powered by Create your own unique website with customizable templates. Need to review Calculating Derivatives that don’t require the Chain Rule? The second is more formal. Step 3: Combine your results from Step 1 2(3x+1) and Step 2 (3). (b) f(x;y) = xy3 + x 2y 2; @f @x = y3 + 2xy2; @f @y = 3xy + 2xy: (c) f(x;y) = x 3y+ ex; @f @x = 3x2y+ ex; @f \end{align*} We could simplify the answer by factoring out the negative signs from the last term, but we prefer to stop there to keep the focus on the Chain rule. A simpler form of the rule states if y – un, then y = nun – 1*u’. ), Solution 2 (more formal). \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= 3\big[\text{stuff}\big]^2 \cdot \dfrac{d}{dx}(\tan x) \\[8px] Here’s a foolproof method: Imagine calculating the value of the function for a particular value of $x$ and identify the steps you would take, because you’ll always automatically start with the inner function and work your way out to the outer function. In this presentation, both the chain rule and implicit differentiation will Let f(x)=6x+3 and g(x)=−2x+5. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function f ∘ g in terms of the derivatives of f and g. The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. Worked example: Derivative of ln(√x) using the chain rule. In this example, the outer function is ex. Now, we just plug in what we have into the chain rule. The derivative of ex is ex, but you’ll rarely see that simple form of e in calculus. Step 1: Differentiate the outer function. d/dy y(½) = (½) y(-½), Step 3: Differentiate y with respect to x. Multiplying 4x3 by ½(x4 – 37)(-½) results in 2x3(x4 – 37)(-½), which when worked out is 2x3/(x4 – 37)(-½) or 2x3/√(x4 – 37). Step 4 Rewrite the equation and simplify, if possible. &= 7(x^2 + 1)^6 \cdot (2x) \quad \cmark \end{align*} Note: You’d never actually write “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Note: keep 3x + 1 in the equation. Need help with a homework or test question? We’ll solve this two ways. This diagram can be expanded for functions of more than one variable, as we shall see very shortly. This section shows how to differentiate the function y = 3x + 12 using the chain rule. Think something like: “The function is some stuff to the power of 3. Since the functions were linear, this example was trivial. Learn More at BYJU’S. The chain rule is a rule for differentiating compositions of functions. The outer function is √, which is also the same as the rational exponent ½. Definition •In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. Example problem: Differentiate y = 2cot x using the chain rule. Chain Rule Example #1 Differentiate $f(x) = (x^2 + 1)^7$. The comment form collects the name and email you enter, and the content, to allow us keep track of the comments placed on the website. Are you working to calculate derivatives using the Chain Rule in Calculus? Step 3. Partial derivative is a method for finding derivatives of multiple variables. This calculus video tutorial shows you how to find the derivative of any function using the power rule, quotient rule, chain rule, and product rule. Solution The outside function is the cosine function: d dx h cos ex4 i = sin ex4 d dx h ex4 i = sin ex4 ex4(4x3): The second step required another use of the chain rule (with outside function the exponen-tial function). Solution: In this example, we use the Product Rule before using the Chain Rule. We’re glad to have helped! = cos(4x)(4). The inner function is the one inside the parentheses: x4 -37. Step 1 (You don’t need us to show you how to do algebra! Compute the integral IS zdrdyd: if D is bounded by the surfaces: D 4. Doing so will give us: f'(x)=5•12(5x-2)^3x Which, when simplified, will give us: f'(x)=60(5x-2)^3x And that is our final answer. We have the outer function $f(u) = u^{99}$ and the inner function $u = g(x) = x^5 + e^x.$ Then $f'(u) = 99u^{98},$ and $g'(x) = 5x^4 + e^x.$ Hence \begin{align*} f'(x) &= 99u^{98} \cdot (5x^4 + e^x) \\[8px] &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*} We could of course simplify the result algebraically to $14x(x^2+1)^2,$ but we’re leaving the result as written to emphasize the Chain rule term $2x$ at the end. &= e^{\sin x} \cdot \left(7x^6 -12x^2 +1 \right) \quad \cmark \end{align*}, Solution 2 (more formal). Applying : (x + 1)½ is the outer function and x + 1 is the inner function. The following chain rule examples show you how to differentiate (find the derivative of) many functions that have an “ inner function ” and an “ outer function.” For an example, take the function y = √ (x 2 – 3). Using the linear properties of the derivative, the chain rule and the double angle formula , we obtain: {y’\left( x \right) }={ {\left( {\cos 2x – 2\sin x} \right)^\prime } } h ' ( x ) = 2 ( ln x ) 1. For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. So the derivative is 7 times that same stuff to the 6th power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] Step 1 Differentiate the outer function, using the table of derivatives. Note: keep 5x2 + 7x – 19 in the equation. We now use the chain rule. Step 3: Differentiate the inner function. Step 2: Differentiate y(1/2) with respect to y. \end{align*}. Include the derivative you figured out in Step 1: Although it’s tedious to write out each separate function, let’s use an extension of the first form of the Chain rule above, now applied to $f\Bigg(g\Big(h(x)\Big)\Bigg)$: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Bigg(g\Big(h(x)\Big)\Bigg) \right]’ &= f’\Bigg(g\Big(h(x)\Big)\Bigg) \cdot g’\Big(h(x)\Big) \cdot h'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the middle function] } \\[5px]&\qquad \times \text{ [derivative of the middle function, evaluated at the inner function]} \\[5px]&\qquad \quad \times \text{ [derivative of the inner function]}\end{align*}}\] The inner function is the one inside the parentheses: x 2 … Your first 30 minutes with a Chegg tutor is free! Chain Rule Practice Problems: Level 01 Chain Rule Practice Problems : Level 02 If 10 men or 12 women take 40 days to complete a piece of work, how long … We won’t write out “stuff” as we did before to use the Chain Rule, and instead will just write down the answer using the same thinking as above: We can view $\left(x^2 + 1 \right)^7$ as $({\text{stuff}})^7$, where $\text{stuff} = x^2 + 1$. Note: keep cotx in the equation, but just ignore the inner function for now. A few are somewhat challenging. Recall that $\dfrac{d}{du}\left(u^n\right) = nu^{n-1}.$ The rule also holds for fractional powers: Differentiate $f(x) = e^{\left(x^7 – 4x^3 + x \right)}.$. Combine the results from Step 1 (2cot x) (ln 2) and Step 2 ((-csc2)). &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*}. We’re glad you found them good for practicing. About "Chain Rule Examples With Solutions" Chain Rule Examples With Solutions : Here we are going to see how we use chain rule in differentiation. The number e (Euler’s number), equivalent to about 2.71828 is a mathematical constant and the base of many natural logarithms. D(5x2 + 7x – 19) = (10x + 7), Step 3. Example 4: Find the derivative of f(x) = ln(sin(x2)). The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. Step 1: Write the function as (x2+1)(½). Chain rule Statement Examples Table of Contents JJ II J I Page5of8 Back Print Version Home Page 21.2.6 Example Find the derivative d dx h cos ex4 i. Solutions. Solutions to Examples on Partial Derivatives 1. For example, let’s say you had the functions: The composition g (f (x)), which is also written as (g ∘ f) (x), would be (x2-3)2. Differentiate ``the square'' first, leaving (3 x +1) unchanged. For instance, $\left(x^2+1\right)^7$ is comprised of the inner function $x^2 + 1$ inside the outer function $(\boxed{\phantom{\cdots}})^7.$ As another example, $e^{\sin x}$ is comprised of the inner function $\sin x$ inside the outer function $e^{\boxed{\phantom{\cdots}}}.$ As yet another example, $\ln{(t^3 – 2t^2 +5)}$ is comprised of the inner function $t^3 – 2t^2 +5$ inside the outer function $\ln(\boxed{\phantom{\cdots}}).$ Since each of these functions is comprised of one function inside of another function — known as a composite function — we must use the Chain rule to find its derivative, as shown in the problems below. Functions that contain multiplied constants (such as y= 9 cos √x where “9” is the multiplied constant) don’t need to be differentiated using the product rule. Differentiating functions that contain e — like e5x2 + 7x-19 — is possible with the chain rule. We have $y = u^7$ and $u = x^2 +1.$ Then $\dfrac{dy}{du} = 7u^6,$ and $\dfrac{du}{dx} = 2x.$ Hence \begin{align*} \dfrac{dy}{dx} &= 7u^6 \cdot 2x \\[8px] x(x2 + 1)(-½) = x/sqrt(x2 + 1). -2cot x(ln 2) (csc2 x), Another way of writing a square root is as an exponent of ½. Identify the mistake(s) in the equation. We’re happy to have helped! &= \left[7\left(x^2 + 1 \right)^6 \cdot (2x) \right](3x – 7)^4 + \left(x^2 + 1 \right)^7 \left[4(3x – 7)^3 \cdot (3) \right] \quad \cmark \end{align*} \] Solution 2 (more formal). Examples az ax ; az ду ; 2. Technically, you can figure out a derivative for any function using that definition. f’ = ½ (x2 – 4x + 2)½ – 1(2x – 4) Step 2: Differentiate the inner function. Just ignore it, for now. \[ \bbox[10px,border:2px solid blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)} }\] Even though few people admit it, almost everyone thinks along the lines of the informal approach in the blue boxes above. The chain rule states formally that . &= e^{\sin x} \cdot \cos x \quad \cmark \end{align*}, Solution 2 (more formal). &= \cos(2x) \cdot 2 \quad \cmark \end{align*}, Solution 2. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Tip: The hardest part of using the general power rule is recognizing when you’re essentially skipping the middle steps of working the definition of the limit and going straight to the solution. We have the outer function $f(z) = \cos z,$ and the middle function $z = g(u) = \tan(u),$ and the inner function $u = h(x) = 3x.$ Then $f'(z) = -\sin z,$ and $g'(u) = \sec^2 u,$ and $h'(x) = 3.$ Hence: \begin{align*} f'(x) &= (-\sin z) \cdot (\sec^2 u) \cdot (3) \\[8px] Solution to Example 1. This is a way of breaking down a complicated function into simpler parts to differentiate it piece by piece. Differentiate f (x) =(6x2 +7x)4 f ( x) = ( 6 x 2 + 7 x) 4 . For example, imagine computing $\left(x^2+1\right)^7$ for $x=3.$ Without thinking about it, you would first calculate $x^2 + 1$ (which equals $3^2 +1 =10$), so that’s the inner function, guaranteed. • Solution 1. No other site explains this nice. Differentiate $f(x) = \left(3x^2 – 4x + 5\right)^8.$. Want access to all of our Calculus problems and solutions? Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] We’ll again solve this two ways. In fact, to differentiate multiplied constants you can ignore the constant while you are differentiating. Chain Rule - Examples. At first glance, differentiating the function y = sin(4x) may look confusing. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²)². This 105. is captured by the third of the four branch diagrams on the previous page. The following equation for h ' (x) comes from applying the chain rule incorrectly. (10x + 7) e5x2 + 7x – 19. Chain Rule: The General Power Rule The general power rule is a special case of the chain rule. We have the outer function $f(u) = u^{-2}$ and the inner function $u = g(x) = \cos x – \sin x.$ Then $f'(u) = -2u^{-3},$ and $g'(x) = -\sin x – \cos x.$ (Recall that $(\cos x)’ = -\sin x,$ and $(\sin x)’ = \cos x.$) Hence \begin{align*} f'(x) &= -2u^{-3} \cdot (-\sin x – \cos x) \\[8px] Huge thumbs up, Thank you, Hemang! 5x2 + 7x – 19. \(g\left( t \right) = {\left( {4{t^2} - 3t + 2} \right)^{ - 2}}\) Solution. &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x)\quad \cmark \\[8px] Differentiate algebraic and trigonometric equations, rate of change, stationary points, nature, curve sketching, and equation of tangent in Higher Maths. cot x. Example question: What is the derivative of y = √(x2 – 4x + 2)? &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x) \quad \cmark \end{align*} We could simplify the answer by factoring out the negative signs from the last term, but we prefer to stop there to keep the focus on the Chain rule. • Solution 1. This imaginary computational process works every time to identify correctly what the inner and outer functions are. D(tan √x) = sec2 √x, Step 2 Differentiate the inner function, which is Get notified when there is new free material. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Want to skip the Summary? Chain rule for partial derivatives of functions in several variables. Example: Chain rule for f(x,y) when y is a function of x The heading says it all: we want to know how f(x,y)changeswhenx and y change but there is really only one independent variable, say x,andy is a function of x. (The outer layer is ``the square'' and the inner layer is (3 x +1). Let u = cosx so that y = u2 It follows that du dx = −sinx dy du = 2u Then dy dx = dy du × du dx = 2u× −sinx = −2cosxsinx Example Suppose we wish to differentiate y = (2x− 5)10. &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] Once you’ve performed a few of these differentiations, you’ll get to recognize those functions that use this particular rule. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . D(sin(4x)) = cos(4x). However, the technique can be applied to any similar function with a sine, cosine or tangent. Let u = 5x - 2 and f (u) = 4 cos u, hence. 1. √ X + 1  7 (sec2√x) ((½) X – ½) = Practice: Chain rule intro. Step 1: Identify the inner and outer functions. &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] More commonly, you’ll see e raised to a polynomial or other more complicated function. We have the outer function $f(u) = \sqrt{u}$ and the inner function $u = g(x) = x^2 + 1.$ Then $\left(\sqrt{u} \right)’ = \dfrac{1}{2}\dfrac{1}{ \sqrt{u}},$ and $\left(x^2 + 1 \right)’ = 2x.$ Hence \begin{align*} f'(x) &= \dfrac{1}{2}\dfrac{1}{ \sqrt{u}} \cdot 2x \\[8px] In this example, the negative sign is inside the second set of parentheses. chain rule example problems MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on CHAIN RULE with easy and logical explanations. The results are then combined to give the final result as follows: Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(… That’s why mathematicians developed a series of shortcuts, or rules for derivatives, like the general power rule. • Solution 3. Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line). (a) f(x;y) = 3x+ 4y; @f @x = 3; @f @y = 4. This exponent behaves the same way as an integer exponent under differentiation – it is reduced by 1 to -½ and the term is multiplied by ½. Therefore sqrt(x) differentiates as follows: Hyperbolic Functions And Their Derivatives. Combine the results from Step 1 (e5x2 + 7x – 19) and Step 2 (10x + 7). Step 1: Rewrite the square root to the power of ½: To differentiate a more complicated square root function in calculus, use the chain rule. Solution 1 (quick, the way most people reason). Worked example: Derivative of √(3x²-x) using the chain rule. &= 8\left(3x^2 – 4x + 5\right)^7 \cdot (6x-4) \quad \cmark \end{align*}. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= -2(\text{stuff})^{-3} \cdot \dfrac{d}{dx}(\cos x – \sin x) \\[8px] In integration, the counterpart to the chain rule is the substitution rule . Note that I’m using D here to indicate taking the derivative. The key is to look for an inner function and an outer function. Sample problem: Differentiate y = 7 tan √x using the chain rule. Note: keep 4x in the equation but ignore it, for now. We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. Combine the results from Step 1 (sec2 √x) and Step 2 ((½) X – ½). If 30 men can build a wall 56 meters long in 5 days, what length of a similar wall can be built … CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. Example: Find the derivative of . 7 (sec2√x) ((½) 1/X½) = 7 (sec2√x) ((1/2) X – ½). We have the outer function $f(u) = u^3$ and the inner function $u = g(x) = \tan x.$ Then $f'(u) = 3u^2,$ and $g'(x) = \sec^2 x.$ (Recall that $(\tan x)’ = \sec^2 x.$) Hence \begin{align*} f'(x) &= 3u^2 \cdot (\sec^2 x) \\[8px] So the derivative is $-2$ times that same stuff to the $-3$ power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] Then. • Solution 2. D(3x + 1)2 = 2(3x + 1)2-1 = 2(3x + 1). The derivative of sin is cos, so: : ), Thank you. \text{Then}\phantom{f(x)= }\\ \frac{df}{dx} &= 7(\text{stuff})^6 \cdot \left(\frac{d}{dx}(x^2 + 1)\right) \\[8px] D(√x) = (1/2) X-½. In this example, the inner function is 4x. \begin{align*} f(x) &= (\text{stuff})^7; \quad \text{stuff} = x^2 + 1 \\[12px] \begin{align*} f(x) &= (\text{stuff})^{-2}; \quad \text{stuff} = \cos x – \sin x \\[12px] Question 1 : Differentiate f(x) = x / √(7 - 3x) Solution : u = x. u' = 1. v = √(7 - 3x) v' = 1/2 √(7 - 3x)(-3) ==> -3/2 √(7 - 3x)==>-3/2 √(7 - 3x) The derivative of cot x is -csc2, so: D(e5x2 + 7x – 19) = e5x2 + 7x – 19. Solution 4: Here we have a composition of three functions and while there is a version of the Chain Rule that will deal with this situation, it can be easier to just use the ordinary Chain Rule twice, and that is what we will do here. ( 1 – ½ ) into simpler parts to differentiate it piece by piece ) – 0, which also. 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