# How do you differentiate sinx+sinxcosx?

Jan 31, 2017

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

#### Explanation:

The derivative of $\sin x$ is $\cos x$.

For $\sin x \cos x$, we need to use the product rule, so let $a , b$ be functions of $x$. Then $\left(a b\right) ' = a ' b + a b '$.

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

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

The final derivative is $\cos x + {\cos}^{2} x - {\sin}^{2} x$.

Jan 31, 2017

$\cos x + \cos 2 x$

#### Explanation:

We require to know the following $\textcolor{b l u e}{\text{standard derivatives}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x , \frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

Also the $\textcolor{b l u e}{\text{trigonometric identity}}$

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

$\sin x \cos x \text{ has to be differentiated using the"color(blue)" product rule}$

$\text{Given "f(x)=g(x).h(x)" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

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

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

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\sin x \cos x\right) = \sin x \left(- \sin x\right) + \cos x \left(\cos x\right)$

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

$\text{Thus } \frac{d}{\mathrm{dx}} \left(\sin x + \sin x \cos x\right) = \cos x + \cos 2 x$