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

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Answer 1 · 2016-02-20T06:25:00

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

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!

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