# How do you find the derivative of f(x) = (sin x)(cos x)?

Jul 21, 2016

f'(x) = cos2x

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{product rule}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f \left(x\right) = g \left(x\right) h \left(x\right) \Rightarrow f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x \text{ and } \frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here $g \left(x\right) = \sin x \Rightarrow g ' \left(x\right) = \cos x$

and $h \left(x\right) = \cos x \Rightarrow h ' \left(x\right) = - \sin x$

Substitute these values into f'(x)

$\Rightarrow f ' \left(x\right) = \cos x \left(- \sin x\right) + \cos x \left(\cos x\right)$

$= {\cos}^{2} x - {\sin}^{2} x$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\cos 2 x = {\cos}^{2} x - {\sin}^{2} x} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow f ' \left(x\right) = {\cos}^{2} x - {\sin}^{2} x = \cos 2 x$
$\textcolor{m a \ge n t a}{\text{----------------------------------------------------------}}$

Alternatively

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin 2 x = 2 \sin x \cos x} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and recognise that $f \left(x\right) = \sin x \cos x = \frac{1}{2} \sin 2 x$

differentiate f(x) using the $\textcolor{b l u e}{\text{chain rule}}$

$\Rightarrow f ' \left(x\right) = \frac{1}{2} \cos 2 x . \frac{d}{\mathrm{dx}} \left(2 x\right) = \frac{1}{2} \cos 2 x .2 = \cos 2 x$

Jul 21, 2016

$\frac{d f \left(x\right)}{d x} \left(\sin x\right) \left(\cos x\right) = 1 - 2 {\sin}^{2} x = \cos 2 x$

#### Explanation:

(d f(x))/(d x) (sin x)( cos x)=?

$\frac{d}{d x} \sin x = \cos x$

$\frac{d}{d x} \cos x = - \sin x$

$y = a \cdot b \text{ ; } {y}^{'} = {a}^{'} \cdot b + {b}^{'} \cdot a$

$\frac{d f \left(x\right)}{d x} \left(\sin x\right) \left(\cos x\right) = \cos x \cdot \cos x - \sin x \cdot \sin x$

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

$\text{remember that :} {\cos}^{2} x = 1 - {\sin}^{2} x$

$\frac{d f \left(x\right)}{d x} \left(\sin x\right) \left(\cos x\right) = 1 - {\sin}^{2} x - {\sin}^{2} x$

$\frac{d f \left(x\right)}{d x} \left(\sin x\right) \left(\cos x\right) = 1 - 2 {\sin}^{2} x = \cos 2 x$