# 3.3 Chain Ruleap Calculus

- 3.3 Chain Ruleap Calculus 14th Edition
- 3.3 Chain Ruleap Calculus 2nd Edition
- 3.3 Chain Ruleap Calculus Solver

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.

Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. All functions are functions of real numbers that return real values. Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). It helps to differentiate composite functions.

- 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.
- This video explains about Topic 3.3 for subject BUM2123 Applied Calculus which will be used in Substitute Blended Learning.
- In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule.

## 3.3 Chain Ruleap Calculus 14th Edition

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.

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)

**Ex 3.3.4**$displaystyle {sqrt{625-x^2}over x^{20}}$(answer)

## 3.3 Chain Ruleap Calculus 2nd Edition

**Ex 3.3.5**Use 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)

## 3.3 Chain Ruleap Calculus Solver

**Ex 3.3.6**Suppose 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 3.3.7**State 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 3.3.8**The **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 $n=5$.