How do you find (dy)/(dx) given siny=x^2?

Jan 24, 2017

$y ' \left(x\right) = \frac{2 x}{\sqrt{1 - {x}^{4}}}$

Explanation:

Differentiate the equation with respect to $x$:

$\sin y = {x}^{2}$

$y ' \cos y = 2 x$

$y ' = \frac{2 x}{\cos} y$

If we restrict the function to the intervals $x \in \left[0 , 1\right]$ and $y \in \left[0 , \frac{\pi}{2}\right]$:

$\cos y = \sqrt{1 - {\sin}^{2} y} = \sqrt{1 - {x}^{4}}$

so that:

$y ' \left(x\right) = \frac{2 x}{\sqrt{1 - {x}^{4}}}$