# What is f'(-pi/3) when you are given f(x)=sin^7(x)?

Mar 27, 2015

It is $\frac{7 \sqrt{3}}{2} ^ 7 = \frac{7 \sqrt{3}}{128}$

Method

$f \left(x\right) = {\sin}^{7} \left(x\right)$

It is very useful to re-write this as $f \left(x\right) = {\left(\sin \left(x\right)\right)}^{7}$ because this makes it clear that what we have is a ${7}^{t h}$ power function.

Use the power rule and the chain rule (This combination is often called the generalized power rule.)

For $f \left(x\right) = {\left(g \left(x\right)\right)}^{n}$, the derivative is $f ' \left(x\right) = n {\left(g \left(x\right)\right)}^{n - 1} \cdot g ' \left(x\right)$,

In other notation $\frac{d}{\mathrm{dx}} \left({u}^{n}\right) = n {u}^{n - 1} \frac{\mathrm{du}}{\mathrm{dx}}$

In either case, for your question $f ' \left(x\right) = 7 {\left(\sin \left(x\right)\right)}^{6} \cdot \cos \left(x\right)$

You could write $f ' \left(x\right) = 7 {\sin}^{6} \left(x\right) \cdot \cos \left(x\right)$

At $x = - \frac{\pi}{3}$, we have
$f ' \left(- \frac{\pi}{3}\right) = 7 {\sin}^{6} \left(- \frac{\pi}{3}\right) \cdot \cos \left(- \frac{\pi}{3}\right) = 7 {\left(\frac{1}{2}\right)}^{6} \left(\frac{\sqrt{3}}{2}\right) = \frac{7 \sqrt{3}}{2} ^ 7$

Mar 27, 2015

$\text{let } y = f \left(x\right)$ $\implies \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right)$

$\implies y = {\sin}^{7} \left(x\right)$

$\text{let } u = \sin \left(x\right) \implies y = {u}^{7}$

$\frac{\mathrm{du}}{\mathrm{dx}} = \cos \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{du}} = 7 \cdot {u}^{6}$

Now, $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$
$= \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ {Do you agree?}
$= 7 {u}^{6} \cdot \cos x$
but remember $u = \sin \left(x\right)$
$\implies f ' \left(x\right) = 7 {\sin}^{6} \left(x\right) \cos \left(x\right)$

$\implies f ' \left(- \frac{\pi}{3}\right) = 7 \cdot {\left(\sin \left(- \frac{\pi}{3}\right)\right)}^{6} \ast \cos \left(- \frac{\pi}{3}\right)$

$= 7 {\left(- \frac{\sqrt{3}}{2}\right)}^{6} \ast \left(\frac{1}{2}\right)$

You Have The Honor To Simplify

NOTE:
{
wondering why im doing all this "let stuff" ?

the reason is there are more than one function in $f \left(x\right)$

** there's : ${\sin}^{7} \left(x\right)$ and there's $\sin \left(x\right)$!!

so to find the $f ' \left(x\right)$ i need to find the $f '$ of ${\sin}^{7} \left(x\right)$
AND the $f '$ of $\sin \left(x\right)$

that's why i need to let $y = f \left(x\right)$
then let $u = \sin \left(x\right)$
}