Question

use the chain rule to find the derivative: y=1/cot^5(x-2)

Answer 1

the right format:
\[y=\frac{ 1 }{ \cot⁵(x⁻²) }\]

Answer 2

so I can get functions f(x)=1/x, g(x)=cot⁵(x⁻²), h(x)=x⁻²

Answer 3

so dy/dx=f'(g(h(x)))*g'(h(x))*h'(x)

Answer 4

uh, what's the derivative of cot⁵? is the derivative of cot x=-csc²x, then is it, i dunno, -csc¹⁰x?

Answer 5

Before differentiating, simplify:
\(f(x)=\dfrac{1}{\cot^5(x-2)}=\tan^5(x-2)\).

This is because cotx=cosx/sinx and tanx =sinx/cosx.

Now you need to know the derivative of tanx. You have two options:
\[(\tan x)'= \tan^2 x+1 = \frac{1}{\cos^2 x}\]
(I don't think you are asked to prove this in this problem, so you can use this).

Remember: \(\tan^5(x-2)=(\tan(x-2))^5\), so if you need the derivative, you must first differentiate the 5th power and then multiply with the derivative of tan(x-2):

\(5\tan^4(x-2) \cdot (\tan (x-2))'=...\)

Answer 6

sorry there was a typo in the first post. It should have been\[y=\frac{ 1 }{ \cot⁵(x⁻²) }\]
So should I pull another function out of the original? as in\[dy/dx=5*(\tan(x⁻²))⁴*\frac{ 1 }{ \cos²(x⁻²) }*\frac{ -2 }{ x³ }\]

Answer 7

Yes, but the derivative of \(x^{-2}\) is \(-2x^{-3}=-\dfrac{2}{x^3}\), so the derivative of your function is \(\dfrac{dy}{dx}=5\tan^4(x^{-2})\cdot \dfrac{1}{\cos^2(x^{-2})}\cdot-\dfrac{2}{x^3}\).

I will try to make it look a little friendlier:

\(\dfrac{dy}{dx}=-\dfrac{10\tan^4(x^{-2})}{x^3\cos^2(x^{-2})}\). Still not pretty, but a little shorter :)

Answer 8

thanks @zehans

Answer 9

YW!