# What is the derivative of sin(x^2y^2)?

Mar 24, 2015

If you want the partial derivatives of $f \left(x , y\right) = \sin \left({x}^{2} {y}^{2}\right)$, they are:

${f}_{x} \left(x , y\right) = 2 x {y}^{2} \cos \left({x}^{2} {y}^{2}\right)$ and

${f}_{y} \left(x , y\right) = 2 {x}^{2} y \cos \left({x}^{2} {y}^{2}\right)$.

If we are considering $y$ to be a function of $x$ and looking for $\frac{d}{\mathrm{dx}} \left(\sin \left({x}^{2} {y}^{2}\right)\right)$, the the answer is:
$\frac{d}{\mathrm{dx}} \left(\sin \left({x}^{2} {y}^{2}\right)\right) = \left[2 x {y}^{2} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}\right] \cos \left({x}^{2} {y}^{2}\right)$
$\frac{d}{\mathrm{dx}} \left(\sin \left({x}^{2} {y}^{2}\right)\right) = \left[\cos \left({x}^{2} {y}^{2}\right)\right] \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} {y}^{2}\right)$
$= = \left[\cos \left({x}^{2} {y}^{2}\right)\right] \cdot \left[2 x {y}^{2} + {x}^{2} 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right]$
$= \left[2 x {y}^{2} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}\right] \cos \left({x}^{2} {y}^{2}\right)$