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Diffstat (limited to 'assignment2.tex')
| -rw-r--r-- | assignment2.tex | 22 |
1 files changed, 11 insertions, 11 deletions
diff --git a/assignment2.tex b/assignment2.tex index fa70c40..f5864bc 100644 --- a/assignment2.tex +++ b/assignment2.tex @@ -84,25 +84,25 @@ \end{enumerate} \item \begin{enumerate} - \item $f'(x) = 4x^3-6x^2.$ - \item $f'(x) = \frac{(x-7)(2x)-(x^2+5)(1)}{(x-7)^2} = \frac{x+2}{x-7}.$ - \item \begin{align*} + \item Using basic rules, $f'(x) = 4x^3-6x^2$. + \item Using the division rule, $f'(x) = \frac{(x-7)(2x)-(x^2+5)(1)}{(x-7)^2} = \frac{x+2}{x-7}$. + \item Using the product rule, \begin{align*} f'(x) &= \left(\sin{\sqrt x}\cdot\sin{\sqrt x}\right) \\ &= 2\cdot\sin\sqrt x\cdot\left(\sin\sqrt x\right)' \\ &= 2\cdot\sin\sqrt x\cdot\cos\sqrt x\cdot\frac12\cdot x^{-\frac12} \\ &= \frac12\sin\left(2\sqrt x\right)\cdot\frac1{\sqrt x} \\ &= \frac{\sin\left(2\sqrt x\right)}{2\sqrt x}. \end{align*} - \item %todo - \item $f'(x) = \left(e^{\tan x}\right)' = e(\tan x)\cdot\tan'x = \frac{e(\tan x)}{\cos^2x}$. - \item $f'(x) = -\frac1{\cos x}\cdot-\sin x=\frac{\sin x}{\tan x}$. - \item \begin{align*} + \item We have $f(x) = 1-\cos^2\left(\sqrt x\right) = \sin^2\left(\sqrt x\right)$, so this is the same as 4c. + \item Using the chain rule, $f'(x) = \left(e^{\tan x}\right)' = e^{\tan x}\cdot\tan'x = \frac{e^{\tan x}}{\cos^2x}$. + \item Using the chain rule, $f'(x) = -\frac1{\cos x}\cdot-\sin x=\tan x$. + \item Using the chain rule, \begin{align*} f'(x) &= \frac1{\sqrt{1-(1-2x)^2}}\cdot-2 \\ - &= -2\cdot\frac1{\sqrt{4x^2+4x}} \\ - &= -2\cdot\frac1{2\sqrt{x^2+x}} \\ - &= -\left(x^2+x\right)^{-\frac12}. + &= -\frac2{\sqrt{-4x^2+4x}} \\ + &= -\frac2{2\sqrt{-x^2+x}} \\ + &= -\left(-x^2+x\right)^{-\frac12}. \end{align*} - \item %todo + \item Using the chain rule, $f'(x) = 10^{x^2}\cdot\ln10\cdot2x$. \end{enumerate} \item % todo |