Stewart – Calculus – 3.2 – Differentiation Rules

September 28, 2009 Clifton Leave a comment

httpv://www.youtube.com/watch?v=4N4qW67m4qo

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 $(1,\frac{1}{2}e)$ .

$\frac{d}{\mathit{dx}}(c)=?<br />$ $\frac{d}{\mathit{dx}}(x^{n})=?<br />$ $\frac{d}{\mathit{dx}}(e^{x})=?<br />$ $(\mathit{cf})'=?<br />$ $(f+g)'=?<br />$ $(f-g)'=?<br />$ $(\mathit{fg})'=?<br />$ $(\frac{f}{g})'=?<br />$

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.

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