How do you take the derivative of $tan^-1(x^2)$ ?

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Answer 1 · 2015-08-26T10:26:54

$y^' = (2x)/(1 + x^4)$

You can differentiate a function $y = tan^(-1)(x^2)$ by using implicit differentiation.

So, if you have a function $y = tan^(-1)(x^2)$ , then you know that you can write

$tan(y) = x^2$

Differentiate both sides with respect to $x$ to get

$d/(dy)(tany) * (dy)/dx = d/dx(x^2)$

$sec^2y * (dy)/dx = 2x$

This is equivalent to saying that

$(dy)/dx = (2x)/sec^2y$

Remember that you have

$color(blue)(sec^2x = 1 + tan^2x)$

which means that you get

$(dy)/dx = (2x)/(1 + tan^2y)$

Finally, replace $tan^2y$ with $x^2$ to get

$(dy)/dx = color(green)((2x)/(1 + x^4))$

How do you take the derivative of tan^-1(x^2)tan^-1(x^2)?