derivative 1

• Oct 28th 2009, 10:37 AM
haebinpark
derivative 1
y= tan^3 (sqrt(4 - x^4))

i did
y = [tan(sqrt4 - x^4)]^3
y'= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[(4-x^4)^-1/2]'
= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2](4x^3)
= 12x^3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2]

do i have to do something with [1/2(4-x^4)^-1/2] ?
• Oct 28th 2009, 12:07 PM
Quote:

Originally Posted by haebinpark
y= tan^3 (sqrt(4 - x^4))

i did
y = [tan(sqrt4 - x^4)]^3
y'= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[(4-x^4)^-1/2]'
= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2](4x^3)
= 12x^3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2]

That isn't the correct derivative.

do i have to do something with [1/2(4-x^4)^-1/2] ?

$\displaystyle y'=3u^2\frac{du}{dx}$ where $\displaystyle u=tan(\sqrt{4-x^4})$

$\displaystyle \frac{du}{dx}=sec^2(\sqrt{4-x^4})\frac{dv}{dx}$ where $\displaystyle v=\sqrt{4-x^4}$

$\displaystyle \frac{dv}{dx}=\frac{4x^3}{2\sqrt{4-x^4}}=\frac{2x^3}{\sqrt{4-x^4}}$

• Oct 28th 2009, 12:23 PM
VonNemo19
Quote:

Originally Posted by haebinpark
y= tan^3 (sqrt(4 - x^4))

i did
y = [tan(sqrt4 - x^4)]^3
y'= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[(4-x^4)^-1/2]'
= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2](4x^3)
= 12x^3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2]

$\displaystyle \frac{d}{dx}\left[\tan^3\sqrt{4-x^4}\right]=3\tan^2\sqrt{4-x^4}\cdot\sec^2\sqrt{4-x^4}\cdot\frac{1}{2}(4-x^4)^{-1/2}(-4x^3)$
$\displaystyle =\frac{-6x^3\tan^2\sqrt{4-x^4}\cdot\sec^2\sqrt{4-x^4}}{\sqrt{4-x^4}}$