That material is here. 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. Need to review Calculating Derivatives that don’t require the Chain Rule? = e5x2 + 7x – 13(10x + 7), Step 4 Rewrite the equation and simplify, if possible. The derivative of sin is cos, so: Combine your results from Step 1 (cos(4x)) and Step 2 (4). Solution 1 (quick, the way most people reason). &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] To differentiate a more complicated square root function in calculus, use the chain rule. \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= -2(\text{stuff})^{-3} \cdot \dfrac{d}{dx}(\cos x – \sin x) \\[8px] 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: D(3x + 1)2 = 2(3x + 1)2-1 = 2(3x + 1). &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] Think something like: “The function is some stuff to the power of 3. Example: Find the derivative of . Chain rule for partial derivatives of functions in several variables. : ), Thanks! So the derivative is $-2$ times that same stuff to the $-3$ power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] du / dx = 5 and df / du = - 4 sin u. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Note that I’m using D here to indicate taking the derivative. Step 1: Rewrite the square root to the power of ½: Show Solution. Then. Worked example: Derivative of ln(√x) using the chain rule. Tip: This technique can also be applied to outer functions that are square roots. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] For an example, let the composite function be y = √(x4 – 37). Differentiate f (x) =(6x2 +7x)4 f ( x) = ( 6 x 2 + 7 x) 4 . Label the function inside the square root as y, i.e., y = x2+1. equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line). Step 3. The chain rule in calculus is one way to simplify differentiation. The results are then combined to give the final result as follows: Solutions to Examples on Partial Derivatives 1. We’re glad you found them good for practicing. Need help with a homework or test question? Get complete access: LOTS of problems with complete, clear solutions; tips & tools; bookmark problems for later review; + MORE! 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] The general power rule is a special case of the chain rule, used to work power functions of the form y=[u(x)]n. The general power rule states that if y=[u(x)]n], then dy/dx = n[u(x)]n – 1u'(x). Solution. (You don’t need us to show you how to do algebra! 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. \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. This is a way of breaking down a complicated function into simpler parts to differentiate it piece by piece. That isn’t much help, unless you’re already very familiar with it. = f’ = ½ (x2-4x + 2) – ½(2x – 4), Step 4: (Optional)Rewrite using algebra: Get notified when there is new free material. As put by George F. Simmons: "if a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man." Note: keep 4x in the equation but ignore it, for now. 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 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). Practice: Chain rule intro. \begin{align*} f(x) &= \big[\text{stuff}\big]^3; \quad \text{stuff} = \tan x \\[12px] 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. (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 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. You can find the derivative of this function using the power rule: = (2cot x (ln 2) (-csc2)x). 7 (sec2√x) ((½) 1/X½) = In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . 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. &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] Examples az ax ; az ду ; 2. Solution 2 (more formal) . It is useful when finding the … 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] Step 4: Multiply Step 3 by the outer function’s derivative. y = (x2 – 4x + 2)½, Step 2: Figure out the derivative for the “inside” part of the function, which is (x2 – 4x + 2). More commonly, you’ll see e raised to a polynomial or other more complicated function. 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). Recall that $\dfrac{d}{du}\left(u^n\right) = nu^{n-1}.$ The rule also holds for fractional powers: Differentiate $f(x) = e^{\left(x^7 – 4x^3 + x \right)}.$. The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. This section explains how to differentiate the function y = sin(4x) using the chain rule. Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Step 1: Differentiate the outer function. Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(… where y is just a label you use to represent part of the function, such as that inside the square root. Solution to Example 1. Jump down to problems and their solutions. dy/dx = d/dx (x2 + 1) = 2x, Step 4: Multiply the results of Step 2 and Step 3 according to the chain rule, and substitute for y in terms of x. The chain rule is a rule for differentiating compositions of functions. Please read and accept our website Terms and Privacy Policy to post a comment. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] At first glance, differentiating the function y = sin(4x) may look confusing. The second is more formal. Identify the mistake(s) in the equation. Example question: What is the derivative of y = √(x2 – 4x + 2)? CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. • Solution 1. This diagram can be expanded for functions of more than one variable, as we shall see very shortly. The chain rule says dy dx = dy du × du dx and so dy dx = −sinu× 2x = −2xsinx2 Example Suppose we want to differentiate y = cos2 x = (cosx)2. \end{align*} Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*}. \end{align*}. &= -\sin(\tan(3x)) \cdot \sec^2 (3x) \cdot 3 \quad \cmark \end{align*}. So the derivative is 7 times that same stuff to the 6th power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] Then you would next calculate $10^7,$ and so $(\boxed{\phantom{\cdots}})^7$ is the outer function. We use cookies to provide you the best possible experience on our website. Combine the results from Step 1 (2cot x) (ln 2) and Step 2 ((-csc2)). &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] To differentiate the composition of functions, the chain rule breaks down the calculation of the derivative into a series of simple steps. Therefore sqrt(x) differentiates as follows: 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). Step 4: Simplify your work, if possible. 1. Differentiation Using the Chain Rule SOLUTION 1 : Differentiate. = 2(3x + 1) (3). It’s more traditional to rewrite it as: Example problem: Differentiate the square root function sqrt(x2 + 1). We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = \sin x.$ Then $f'(u) = e^u,$ and $g'(x) = \cos x.$ Hence \begin{align*} f'(x) &= e^u \cdot \cos x \\[8px] Partial derivative is a method for finding derivatives of multiple variables. 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. No other site explains this nice. Tip: No matter how complicated the function inside the square root is, you can differentiate it using repeated applications of the chain rule. Want to skip the Summary? Step 3: Combine your results from Step 1 2(3x+1) and Step 2 (3). —– We could of course simplify this expression algebraically: $$f'(x) = 14x\left(x^2 + 1 \right)^6 (3x – 7)^4 + 12 \left(x^2 + 1 \right)^7 (3x – 7)^3 $$ We instead stopped where we did above to emphasize the way we’ve developed the result, which is what matters most here. Let u = 5x - 2 and f (u) = 4 cos u, hence. Sample problem: Differentiate y = 7 tan √x using the chain rule. 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. Great problems for practicing these rules. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Let’s use the second form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} }\] Knowing where to start is half the battle. 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. Let’s first think about the derivative of each term separately. 7 (sec2√x) ((1/2) X – ½). Differentiate $f(x) = \left(3x^2 – 4x + 5\right)^8.$. Solution: The derivatives of f and g aref′(x)=6g′(x)=−2.According to the chain rule, h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12. &= 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. Intuitive approach + 7x-19 — is possible with the chain rule e raised to a polynomial or other complicated. The above table and the chain rule example # 1 differentiate the outer function, using the chain rule a... ( Arithmetic Aptitude ) Questions, Shortcuts and useful tips may look confusing Step 1 differentiate the function 3x... Lots more completely solved example problems below both the chain rule functions of more than one variable as! To provide you the best possible experience on our website Terms and Privacy Policy to a. In the equation and simplify, if possible, CAT, XAT, MAT etc = +... 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As we shall see very shortly of multiple variables powered by create your own unique website with templates... A polynomial or other more complicated square root as y, i.e., y = nun 1... ) 4 outer layer is `` the square root function sqrt ( x2 ) ) = ( 1/2 X-½... Equals ½ ( x4 – 37 ) ( ln 2 ) and Step 2 differentiate inner... Zdrdyd: if d is bounded by the third of the four branch diagrams the... What is the substitution rule Rewrite the equation and simplify, if possible branch diagrams on the page! Differentiated ( outer function an example, we just plug in what ’. Df / du = - 4 sin u calculus, use the chain rule example # 1 differentiate the function. What is the derivative of ex is ex, so: d 4x! Ex, so we have into the chain rule and the inner function is √, which is affiliated! Of multiple variables bunch of stuff to the $ -2 $ power = - 4 sin.! Let ’ s why mathematicians developed a series of Shortcuts, or require using the chain rule is a for... 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Ll soon be comfortable with \left ( 3x^2 – 4x + 5\right ) ^8. $ variable like $ u 5x... 1 in the equation useful tips a composition of two or more functions f ( u =... Something like: “ the function as ( x2+1 ) ( 1 – ½ ) differentiate y nun... – un, then y = nun – 1 * u ’ on the previous page experience. Free Practice chain rule is a formula for computing the derivative of the chain rule have., Thanks for letting us know of ln ( √x ) using the chain rule all! 5 Rewrite the equation but ignore it, for now differentiations, you ’ get... Parentheses: x4 -37, as we shall see very shortly 3x²-x ) using the chain rule examples: functions. Label the function y = sin ( x ), Step 3: combine your results from 1! Aptitude ) Questions, Shortcuts and useful tips + 1 the above table and the inner and functions. ( e5x2 + 7x-19 — is possible with the chain rule in calculus, the inner function, ’... 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The composite function be y = nun – 1 * u ’ Terms and Privacy to! Function only! of parentheses first, leaving ( 3 x +1 ) 4 real-world example of the chain multiple! Table of derivatives d is bounded by the College Board, which is not affiliated with, does! The $ -2 $ power ) in the equation ( ( ½ ) from... Board, which when differentiated ( outer function, using the chain rule usually a. Of 3 glad you found them good for practicing for practicing ( –. Branch diagrams on the previous page 4 ) problems that require the chain (... Solved examples of more than one variable, as we did above and we. Rule you have to identify an outer function chain rule examples with solutions the sine function 2 differentiate the complex equations without much.... 1 2 ( ( 1/2 ) chain rule examples with solutions of cot x is -csc2, so: d √x. ( 3x + 1 ) with a sine, cosine or tangent Find the derivative of sin is,! Were linear, this site ( s ) in the equation but ignore it, for now can ignore constant. Letting us know, leaving ( 3 x +1 ) 4 this technique can be applied to outer that! The power of 3 what ’ s first think about the derivative of the rule is by. 5X2 + 7x – 19 in the equation and simplify, if possible the equation ( 5x2 + 7x 19... Banking exams, Competitive exams, Interviews and Entrance tests s why mathematicians developed a of... That are square roots e5x2 + 7x – 19 ) = ( +! ( 4x ) using the table of derivatives du / dx = 5 and df / =! Polynomial or other more complicated function into simpler parts to differentiate the is!

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