How do you differentiate f(x)= sin(x^2)cos(x^2) using the product rule?

Dec 5, 2015

$f ' \left(x\right) = 2 x \cos \left(2 {x}^{2}\right)$

Explanation:

$f ' \left(x\right) = \cos \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left[\sin \left({x}^{2}\right)\right] + \sin \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left[\cos \left({x}^{2}\right)\right]$

$\frac{d}{\mathrm{dx}} \left[\sin \left({x}^{2}\right)\right] = \cos \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] = 2 x \cos \left({x}^{2}\right)$

$\frac{d}{\mathrm{dx}} \left[\cos \left({x}^{2}\right)\right] = - \sin \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] = - 2 x \sin \left({x}^{2}\right)$

Plug back in.

$f ' \left(x\right) = 2 x {\cos}^{2} \left({x}^{2}\right) - 2 x {\sin}^{2} \left({x}^{2}\right)$

$f ' \left(x\right) = 2 x \left({\cos}^{2} \left({x}^{2}\right) - {\sin}^{2} \left({x}^{2}\right)\right)$

Note that ${\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right) = \cos \left(2 a\right)$.

$f ' \left(x\right) = 2 x \cos \left(2 {x}^{2}\right)$