How do you find the derivative of $y = arctan(x^2)$ ?

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Answer 1 · 2016-09-16T03:24:22

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

Note that $d/dxarctan(x)=1/(1+x^2)$ . Thus, according to the chain rule:

$d/dxarctan(f(x))=1/(1+f(x)^2)*f'(x)$

Thus:

$dy/dx=d/dxarctan(x^2)=1/(1+(x^2)^2)*d/dxx^2=(2x)/(1+x^4)$

Answer 2 · 2016-09-16T03:28:14

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

Rearrange the equation:

$tan(y)=x^2$

Differentiate both sides. Recall to use the chain rule on the left hand side.

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

Note that $sec^2(y)=tan^2(y)+1$ :

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

Since $tan(y)=x^2$ :

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

Solving for $dy/dx$ :

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

How do you find the derivative of y = arctan(x^2)y = arctan(x^2)?