# How do you use the product rule to differentiate  sin(x^2)(cos(x^2)) ?

May 14, 2017

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

#### Explanation:

d/dx[sin(x^2)cos(x^2)]=(sin(x^2))(d/dx[cos(x^2)])+(d/dx[sin(x^2)])(cos(x^2))= (sin(x^2))(-2xsin(x^2))+(2xcos(x^2))(cos(x^2))=2xcos^2(x^2)-2xsin^2(x^2)

Don't forget to chain rule ${x}^{2}$

May 14, 2017

$2 x \cdot \cos \left(2 \cdot {x}^{2}\right)$

#### Explanation:

Let $f \left(x\right) = \sin \left({x}^{2}\right)$ and $g \left(x\right) = \cos \left({x}^{2}\right)$

You have :
$f ' \left(x\right) = 2 x \cdot \cos \left({x}^{2}\right)$
and
$g ' \left(x\right) = - 2 x \cdot \sin \left({x}^{2}\right)$

then you apply the formula :
$f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

which gives you :
$2 x \cdot \left({\cos}^{2} \left({x}^{2}\right) - {\sin}^{2} \left({x}^{2}\right)\right)$
Hence, with the formula $\cos \left(2 a\right) = {\cos}^{2} a - {\sin}^{2} a$,
$2 x \cdot \cos \left(2 \cdot {x}^{2}\right)$

May 14, 2017

The differentiation of a product is determined by applying this method:
$\text{ }$
$\frac{d}{\mathrm{dx}} \left(f \left(x\right) \times g \left(x\right)\right) = \frac{d}{\mathrm{dx}} f \left(x\right) \times g \left(x\right) + \frac{d}{\mathrm{dx}} g \left(x\right) \times f \left(x\right)$
$\text{ }$
$\left(\sin \left({x}^{2}\right) \left(\cos \left({x}^{2}\right)\right)\right) '$
$\text{ }$
$\left(\sin \left({x}^{2}\right)\right) ' \times \cos \left({x}^{2}\right) + \left(\cos {x}^{2}\right) ' \times \sin {x}^{2}$
$\text{ }$
$= 2 x \cos {x}^{2} \times \cos {x}^{2} - 2 x \sin {x}^{2} \times \sin {x}^{2}$
$\text{ }$
$= 2 x \left(\cos {x}^{2} \times \cos {x}^{2} - \sin {x}^{2} \times \sin {x}^{2}\right)$
$\text{ }$
$= 2 x \left({\cos}^{2} \left({x}^{2}\right) - {\sin}^{2} \left({x}^{2}\right)\right)$
$\text{ }$
$= 2 x \cos \left(2 {x}^{2}\right)$