Question

How do I find the derivative of `y = arccos((x-3)^2)`?

Answer 1

`dy/dx=d/dx(cos^-1 (x-3)^2)=(-2x+6)/sqrt(1-(x-3)^4)`

Explanation:

The formula to find the derivative of `cos^-1 u` is

`d/dx(cos^-1 u)=-1/sqrt(1-u^2)*d/dx(u)`

So from the given `y=cos^-1 (x-3)^2`
Let `u=(x-3)^2`

from the formula

`dy/dx=d/dx(cos^-1 u)=-1/sqrt(1-u^2)*d/dx(u)`

`dy/dx=d/dx(cos^-1 (x-3)^2)=-1/sqrt(1-((x-3)^2)^2)*d/dx((x-3)^2)`

`dy/dx=d/dx(cos^-1 (x-3)^2)=`

`-1/sqrt(1-(x-3)^4)*2(x-3)*d/dx(x-3)`

`dy/dx=d/dx(cos^-1 (x-3)^2)=(-2(x-3))/sqrt(1-(x-3)^4)*(1)`

`dy/dx=d/dx(cos^-1 (x-3)^2)=(-2x+6)/sqrt(1-(x-3)^4)`

have a nice day!