# What is the derivative of sinxcosx?

##### 3 Answers
Jan 7, 2017

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

#### Explanation:

The product rule can be used to differentiate any function of the form $f \left(x\right) = g \left(x\right) h \left(x\right)$. It states that color(red)(f'(x) = g'(x)h(x) + g(x)h'(x).

The derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $- \sin x$.

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

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

Use the identity $\cos 2 x = {\cos}^{2} x - {\sin}^{2} x$:

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

Hopefully this helps!

Jan 7, 2017

$\cos 2 x$

#### Explanation:

Since this is the product of 2 functions, differentiate using the $\textcolor{b l u e}{\text{product rule}}$

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

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

The following $\textcolor{b l u e}{\text{derivatives}}$ should be known.

color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(sinx)=cosx" and " d/dx(cosx)=-sinx)color(white)(2/2)|)))

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

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

substitute these values into (A)

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

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

This may be simplified using the $\textcolor{b l u e}{\text{trigonometric identity}}$

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

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

Jan 7, 2017

$\cos 2 x .$

#### Explanation:

Let $y = \sin x \cos x = \frac{1}{2} \left(2 \sin x \cos x\right) = \frac{1}{2} \sin 2 x = \frac{1}{2} \sin \phi ,$

where, $\phi = 2 x .$

Here, we first need to know that, $\frac{d}{\mathrm{dt}} \left\{k f \left(t\right)\right\} = k \frac{d}{\mathrm{dt}} \left\{f \left(t\right)\right\} ,$

where, $k$ is a costant.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{1}{2} \sin \phi\right) = \frac{1}{2} \frac{d}{\mathrm{dx}} \left(\sin \phi\right)$

We now use the Chain Rule, which states that :

If $y$ is a function (fun.) of $\phi , \mathmr{and} , \phi \text{ is a fun. of } x$ [ so

that $y$ becomes a fun. of $x$], then, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dp} h i} \frac{\mathrm{dp} h i}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \frac{d}{\mathrm{dx}} \left(\sin \phi\right)$

$= \frac{1}{2} \frac{d}{\mathrm{dp} h i} \left(\sin \phi\right) \left(\frac{\mathrm{dp} h i}{\mathrm{dx}}\right)$

$= \frac{1}{2} \cos \phi \frac{d}{\mathrm{dx}} \left(2 x\right) \ldots \ldots \ldots \ldots \left[\because , \phi = 2 x\right]$

$= \left(\frac{1}{2} \cos \phi\right) \left(2 \frac{d}{\mathrm{dx}} \left(x\right)\right)$

$= \cos \phi = \cos 2 x$

Enjoy Maths!