# Stewart – Calculus – 3.2 – Differentiation Rules

THE PRODUCT RULE $\frac{d}{\mathit{dx}}[f(x)g(x)]=f(x)\frac{d}{\mathit{dx}}[g(x)]+g(x)\frac{d}{\mathit{dx}}[f(x)]$

(a) If $f(x)=xe^{x}$, find $f'(x)$.

(b) Find the nth derivative, $f^{(n)}(x)$.

Differentiate the function $f(t)=\sqrt{t}(a+\mathit{bt})$

If $f(x)=\sqrt{x} \hspace{1 mm} g(x)$, where $g(4)=2$ and $g'(4)=3$, find $f'(4)$.

THE QUOTIENT RULE $\frac{d}{\mathit{dx}}[\frac{f(x)}{g(x)}]=\frac{g(x)\frac{d}{\mathit{dx}}[f(x)]\text{--}f(x)\frac{d}{\mathit{dx}}[g(x)]}{[g(x)]^{2}}$

Let $y=\frac{x^{2}+x\text{--}2}{x^{3}+6}$

Find an equation of the tangent line to the curve $y=\frac{e^{x}}{(1+x^{2})}$ at the point $(1,\frac{1}{2}e)$. $\frac{d}{\mathit{dx}}(c)=?
$ $\frac{d}{\mathit{dx}}(x^{n})=?
$ $\frac{d}{\mathit{dx}}(e^{x})=?
$ $(\mathit{cf})'=?
$ $(f+g)'=?
$ $(f-g)'=?
$ $(\mathit{fg})'=?
$ $(\frac{f}{g})'=?
$  SOLUTION 1
Using the Product Rule SOLUTION 2
If we first use the laws of exponents to rewrite f(t), then we can proceed directly without using the Product Rule.    0       