Homework:  Chain Rule

Tyler Seacrest

30 Homework: Chain Rule

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Chain Rule Definition and example
Take the derivative of the following functions, each of which involves the chain rule.
1. $a (x) = (x^{2} + 5)^{20}$
  
  $40 x (x^{2} + 5)^{19}$
  
  ans
2. $b (x) = e^{x^{2}}$
  
  $2 x e^{x^{2}}$
  
  ans
3. $c (x) = (k x + r)^{n}$ for constants $k$ , $r$ , $n$ .
  
  $k n (k x + r)^{n - 1}$
  
  ans
4. $d (x) = (\ln (x))^{3} + \ln (x^{3})$
  
  $\frac{3 (\ln (x))^{2}}{x} + \frac{3}{x}$
  
  ans
5. $e (x) = \sin (\cos (x))$
  
  $- \sin (x) \cos (\cos (x))$
  
  ans
6. $f (x) = e^{\sin (x) + \cos (x)}$
  
  $(\cos (x) - \sin (x)) e^{\sin (x) + \cos (x)}$
  
  ans
7. $g (x) = \sqrt{3 x^{2} - 5 x + 6}$
  
  $\frac{6 x - 5}{2 \sqrt{3 x^{2} - 5 x + 6}}$
  
  ans
8. $h (x) = e^{- x}$
  
  $- e^{- x}$
  
  ans
For each problem, try simplifying the logarithm first, then taking the derivative.
1. $\frac{d}{d x} \ln (x^{3})$
  
  $\frac{3}{x}$
  
  ans
2. $\frac{d}{d x} \ln (x e^{x})$
  
  $\frac{1}{x} + 1$
  
  ans
Use logarithm rules to explain why $\frac{d}{d x} \ln (e^{5} \cdot x) = \frac{d}{d x} \ln (x)$ .

Using logarithm rules, we have that $l n (e^{5} \cdot x) = \ln (x) + \ln (e^{5}) = \ln (x) + 5$ . This has the same derivative as $\ln (x)$ since we are just adding a constant.

ans
Recall that $\ln (x)$ and $e^{x}$ are inverse functions. This means that $\ln (e^{x}) = x$ , and $e^{\ln (x)} = x$ (that is, the $e$ and the $\ln$ cancel out if you do one right after the other). This fact allows us to compute $\frac{d}{d x} 2^{x}$ .
1. Simplify $e^{\ln (2)}$
  
  $= 2$
  
  ans
2. Simplify $\ln (e^{2})$ .
  
  $= 2$
  
  ans
3. Simplify $e^{\ln (2) + x}$
  
  $2 e^{x}$
  
  ans
4. Simplify $(e^{\ln (2) x})$
  
  $2^{x}$
  
  ans
5. Use part (d) to compute $\frac{d}{d x} 2^{x}$ .
  
  $\ln (2) 2^{x}$
  
  ans

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30 Homework: Chain Rule

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