Calculus is a branch of mathematics that studies rates of change. This first part of a two part tutorial with examples covers the concept of limits, differentiating by first principles, rules of differentiation and applications of differential calculus. Chain Rule of Differentiation in Calculus. The chain rule of differentiation of functions in calculus is presented along with several examples and detailed solutions and comments. Also in this site, Step by Step Calculator to Find Derivatives Using Chain Rule. Calculus: Early Transcendentals 8th Edition answers to Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 204 2 including work step by step written by community members like you. Textbook Authors: Stewart, James, ISBN-10:, ISBN-13: 978-1-28574-155-0, Publisher: Cengage Learning. Find the Derivative f(x)=x(x-4)^3. Differentiate using the chain rule, which states that is where. Tap for more steps. To apply the Chain Rule.
- 3.3 Chain Ruleap Calculus Calculator
- 3.3 Chain Ruleap Calculus Solver
- 3.3 Chain Ruleap Calculus 2nd Edition
- 3.3 Chain Ruleap Calculus 14th Edition
An Approach
to
C A L C U L U S
Table of Contents | Home
7
The derivative of a function of a function
f (x) = x5 and g(x) = x2+ 1.
If we now let g(x) be the argument of f, then f will be a function of g.
f (g(x)) = (x2+ 1)5.
(Topic 3 of Precalculus.)
What is the derivative of f (g(x)) ?
First, note that
| d f(x) dx | = 5x4. |
That is: The derivative of f with respect to its argument (which in this case is x) is equal to 5 times the 4th power of the argument.
This means that if g -- or any variable -- is the argument of f, the same form applies:
| d f(g) dg | = 5g4. |
| d f(h) dh | = 5h4. |
| d f(v) dv | = 5v4. |
In other words, we can really take the derivative of a function of an argument only with respect to that argument.
Therefore, since g = x2+ 1,
| d f(g) dg | = 5g4 | = 5(x2+ 1)4. |
Next, the derivative of g is 2x. What is called the chain rule states the following:
'If f is a function of g and g is a function of x,
then the derivative of f with respect to x
is equal to the derivative of f(g) with respect to g
times the derivative of g(x) with respect to x.'
Therefore according to the chain rule, the derivative of
(x2+ 1)5
is
5(x2+ 1)4· 2x.
Note: In (x2+ 1)5, x2+ 1 is 'inside' the 5th power, which is 'outside.' We take the derivative from outside to inside. When we take the outside derivative, we do not change what is inside. We then multiply by the derivative of what is inside.
To decide which function is outside, decide which you would have to evaluate last.
To evaluate
(x2+ 1)5,
you would first have to evaluate x2+ 1. Then you would take its 5th power. The 5th power therefore is outside. That is why we take that derivative first.
When we write f(g(x)), f is outside g. We take the derivative of f with respect to g first.
| Example 1.f(x) = | . What is its derivative? |
Solution. This has the form f (g(x)). What function is f, that is, what is outside, and what is g, which is inside?
By clicking 'proceed' you consent to sharing your name, email address, membership status, Hogwarts house and wand between Wizarding World Digital LLC and the Harry Potter Shop for the purpose of providing services to you, personalising your experience, to send you customised offers, and for other purposes outlined under our privacy policy. Join the Harry Potter Fan Club for free to discover your Hogwarts house. Get closer to the magic this Christmas with our new collection of festive treats. Harry potter's &daily prophet& newspaper pdf.
g is x4 − 2 because that is inside the square root function, which is f. The derivative of the square root is given in the Example of Lesson 6. For any argument g of the square root function,
Here, g is x4 − 2. Therefore, since the derivative of x4 − 2 is 4x3,
| d dx | = ½(x4 − 2)−½· 4x3 = 2x3(x4 − 2)−½. |
Example 2. What is the derivative of y = sin3x ?
Solution. This is the 3rd power of sin x. To decide which function is outside, how would you evaluate that?
You would first evaluate sin x, and then take its 3rd power. sin x is inside the 3rd power, which is outside.
Now, the derivative of the 3rd power -- of g3 -- is 3g2. Therefore, accepting for the moment that the derivative of sin x is cos x (Lesson 12), the derivative of sin3x -- from outside to inside -- is
3 sin2x· cos x.
| Example 3. What is the derivative of | 1 x3 + 1 | ? |
| Solution. x3 + 1 is inside the function | 1 x | = x−1, |
whose derivative is −x−2 ; (Problem 4, Lesson 4). We have, then,
| 1 x3 + 1 | = | (x3 + 1)−1 | . |
Therefore, its derivative is
3.3 Chain Ruleap Calculus Calculator
−(x3 + 1)−2· 3x2
Example 4.Assume that y is a function of x. y = y(x). Apply the chain rule to
| d dx | y2 | . |
| Solution. | dy2 dx | = | dy2 dy | · | dy dx | = | 2y | dy dx | . |
y, which we are assuming to be a function of x, is inside the function y2. The derivative of y2with respect to y is 2y. As for the derivative of
| y with respect to x, we indicate it as | dy dx | . (See Lesson 5.) |
Problem 1. Calculate the derivative of (x2−3x + 5)9.
To see the answer, pass your mouse over the colored area.
To cover the answer again, click 'Refresh' ('Reload').
Do the problem yourself first!
9(x2−3x + 5)8(2x − 3)
Problem 2. Calculate the derivative of (x4 − 3x2+ 4)2/3.
2/3(x4 − 3x2+ 4)−1/3(4x3 − 6x)
Problem 3. Calculate the derivative of sin5x.
5 sin4x cos x
Problem 4. Calculate the derivative of sin x5.
The inside function is x5 -- you would evaluate that last. The outside function is sin x. (This is the sine of x5.) Therefore, the derivative is
cos x5· 5x4.
Problem 5. Calculate the derivative of sin (1 + 2).
cos (1 + 2)x−1/2.
Problem 6. Calculate the derivative of
¼(sin x)−3/4 cos x.
Example 5. More than two functions. The chain rule can be extended to more than two functions. For example, let
| f(x) | = | . |
The outside function is the square root. Inside that is (1 + a 2nd power). And inside that is sin x.
The derivative therefore is
3.3 Chain Ruleap Calculus Solver
| ½(1 + sin2x)−1/2· 2 sin x· cos x | = | sin x cos x | . |
| Problem 7. Calculate the derivative of |
(Compare Example 3.)
| −[sin (x2+ 5)]−2· cos (x2+ 5)· 2x | = | − | 2x cos (x2+ 5) sin2(x2+ 5) |
| Problem 8. Calculate the derivative of |
Problem 9. Assume that y is a function of x, and apply the chain rule to express each derivative with respect to x.
| a) | d dx | y3 = | 3y2 | dy dx |
| b) | d dx | sin y = | cos y | dy dx |
| c) | d dx | = | ½y−½ | dy dx |
Proof of the chain rule
To prove the chain rule let us go back to basics. Let f be a function of g, which in turn is a function of x, so that we have f(g(x)). Then when the value of g changes by an amount Δg, the value of f will change by an amount Δf. We will have the ratio
| Δf Δg | . |
Again, since g is a function of x, then when x changes by an amount Δx, g will change by an amount Δg. We will have the ratio
| Δg Δx | . |
But the change in x affects f because it depends on g. We will have
| Δf Δx | . It will be the product of those ratios: |
| Δf Δx | = | Δf Δg | · | Δg Δx | . |
Let us now take the limit as Δx approaches 0. Then the change in g(x) -- Δg -- will also approach 0. Therefore, since the limit of a product is equal to the product of the limits (Lesson 2), and by definition of the derivative:
This is the chain rule.
Table of Contents | Home
Please make a donation to keep TheMathPage online.
Even $1 will help.
Copyright © 2020 Lawrence Spector
Questions or comments?
E-mail:themathpage@yandex.com
Consider the product of two simple functions, say$ds f(x)=(x^2+1)(x^3-3x)$. An obvious guess for the derivative of $f$ isthe product of the derivatives of the constituent functions:$ds (2x)(3x^2-3)=6x^3-6x$. Is this correct? We can easily check, byrewriting $f$ and doing the calculation in a way that is known towork. First, $ds f(x)=x^5-3x^3+x^3-3x=x^5-2x^3-3x$, and then$ds f'(x)=5x^4-6x^2-3$. Not even close! What went 'wrong'? Well,nothing really, except the guess was wrong.
So the derivative of $f(x)g(x)$ is NOT as simple as$f'(x)g'(x)$. Surely there is some rule for such a situation? Thereis, and it is instructive to 'discover' it by trying to do thegeneral calculation even without knowing the answer in advance.$$eqalign{{dover dx}(&f(x)g(x)) = lim_{Delta x to0} {f(x+Delta x)g(x+Delta x) - f(x)g(x)over Delta x}cr&=lim_{Delta x to0} {f(x+Delta x)g(x+Delta x)-f(x+Delta x)g(x) + f(x+Delta x)g(x)- f(x)g(x)over Delta x}cr &=lim_{Delta x to0} {f(x+Delta x)g(x+Delta x)-f(x+Delta x)g(x)over Delta x} + lim_{Delta x to0} {f(x+Delta x)g(x)- f(x)g(x)over Delta x}cr &=lim_{Delta x to0} f(x+Delta x){ g(x+Delta x)-g(x)over Delta x} + lim_{Delta x to0} {f(x+Delta x)- f(x)over Delta x}g(x)cr &=f(x)g'(x) + f'(x)g(x)cr}$$A couple of items here need discussion. First, we used a standardtrick, 'add and subtract the same thing', to transform what we hadinto a more useful form. After some rewriting, we realize that we havetwo limits that produce $f'(x)$ and $g'(x)$. Of course, $f'(x)$ and$g'(x)$ must actually exist for this to make sense.We also replaced$ds lim_{Delta xto0}f(x+Delta x)$ with $f(x)$—why is this justified?
What we really need to know here is that $ds lim_{Delta xto 0}f(x+Delta x)=f(x)$, or in the language ofsection 2.5, that $f$ is continuousat $x$. We already know that $f'(x)$ exists (or the whole approach,writing the derivative of $fg$ in terms of $f'$ and $g'$, doesn't makesense). This turns out to imply that $f$ is continuous as well. Here'swhy:$$eqalign{lim_{Delta xto 0} f(x+Delta x) &= lim_{Delta xto 0} (f(x+Deltax) -f(x) + f(x))cr&= lim_{Delta xto 0} {f(x+Delta x) -f(x)over Delta x}Delta x +lim_{Delta xto 0} f(x)cr&=f'(x)cdot 0 + f(x) = f(x)cr}$$
To summarize: the product rule says that$${dover dx}(f(x)g(x)) = f(x)g'(x) + f'(x)g(x).$$
Returning to the example we started with, let $ds f(x)=(x^2+1)(x^3-3x)$.Then $ds f'(x)=(x^2+1)(3x^2-3)+(2x)(x^3-3x)=3x^4-3x^2+3x^2-3+2x^4-6x^2=5x^4-6x^2-3$, as before. In this case it is probably simpler tomultiply $f(x)$ out first, then compute the derivative; here's anexample for which we really need the product rule.
3.3 Chain Ruleap Calculus 2nd Edition
Example 3.3.1 Compute the derivative of $ds f(x)=x^2sqrt{625-x^2}$. We havealready computed $ds {dover dx}sqrt{625-x^2}={-xoversqrt{625-x^2}}$. Now$$f'(x)=x^2{-xoversqrt{625-x^2}}+2xsqrt{625-x^2}={-x^3+2x(625-x^2)over sqrt{625-x^2}}={-3x^3+1250xover sqrt{625-x^2}}.$$
Exercises 3.3
In 1–4, find the derivatives of the functions using the product rule.
Ex 3.3.1$ds x^3(x^3-5x+10)$(answer)
Ex 3.3.2$ds (x^2+5x-3)(x^5-6x^3+3x^2-7x+1)$(answer)
Ex 3.3.3$ds sqrt{x}sqrt{625-x^2}$(answer)
3.3 Chain Ruleap Calculus 14th Edition
Ex 3.3.4$displaystyle {sqrt{625-x^2}over x^{20}}$(answer)
Ex 3.3.5Use the product rule to compute the derivative of $ds f(x)=(2x-3)^2$. Sketch the function. Find an equation of the tangent line to the curve at $x=2$. Sketch the tangent line at $x=2$.(answer)
Ex 3.3.6Suppose that $f$, $g$, and $h$ are differentiable functions.Show that $(fgh)'(x) = f'(x) g(x)h(x) + f(x)g'(x) h(x) + f(x) g(x)h'(x)$.
Ex 3.3.7State and prove a rule to compute $(fghi)'(x)$, similar to the rule in the previous problem.
Remark 3.3.2 {Product notation}Suppose $ds f_1 , f_2 , ldots f_n$ are functions.The product of all these functions can be written$$ prod _{k=1 } ^n f_k.$$This is similar to the use of $ds sum$ to denote a sum.For example,$$prod _{k=1 } ^5 f_k =f_1 f_2 f_3 f_4 f_5$$and$$prod _ {k=1 } ^n k = 1cdot 2 cdot ldots cdot n = n!.$$We sometimes use somewhat more complicated conditions; for example$$prod _{k=1 , kneq j } ^n f_k$$denotes the product of $ds f_1$ through $ds f_n$ except for $ds f_j$.For example, $$prod _{k=1 , kneq 4} ^5 x^k = xcdot x^2 cdot x^3 cdot x^5 =x^{11}.$$
Ex 3.3.8The generalized product rule says that if $ds f_1 , f_2 ,ldots ,f_n$ are differentiable functions at $x$ then$${dover dx}prod _{k=1 } ^n f_k(x) = sum _{j=1 } ^n left(f'_j (x) prod _{k=1 , kneq j} ^n f_k (x)right).$$Verify that this is the same as your answer to the previous problemwhen $n=4$,and write out what this says when $n=5$.