nemired9

2021-12-26

What is the derivative of $h\left(x\right)=\mathrm{sin}2x\mathrm{cos}2x$

Orlando Paz

Expert

Step 1
Product Rule of differentiation:
If f(x) and g(x) are any two differentiable functions, then
${f\left(g\right)}^{\prime }\left(x\right)=f\left(x\right)×{g}^{\prime }\left(x\right)+{f}^{\prime }\left(x\right)×g\left(x\right)$
Chain rule of differentiation:
$\frac{d}{dx}f\left(g\left(x\right)\right)={f}^{\prime }\left(g\left(x\right)\right)×\frac{d}{dx}g\left(x\right)$
$={f}^{\prime }\left(g\left(x\right)\right)×{g}^{\prime }\left(x\right)×\frac{d}{dx}\left(x\right)$
$={f}^{\prime }\left(g\left(x\right)\right)×{g}^{\prime }\left(x\right)$
The given function is $h\left(x\right)=\mathrm{sin}2x\mathrm{cos}2x$
Step 2
Differentiate the given function using the product rule as follows.
${h}^{\prime }\left(x\right)=\frac{d}{dx}\left(\mathrm{sin}2x\mathrm{cos}2x\right)$
$=\mathrm{sin}2x\frac{d}{dx}\left(\mathrm{cos}2x\right)+\mathrm{cos}2x\frac{d}{dx}\left(\mathrm{sin}2x\right)$
(U sin g product rule)
$=\mathrm{sin}2x\left(-\mathrm{sin}2x\right)\left(2\right)+\mathrm{cos}2x\left(\mathrm{cos}2x\right)\left(2\right)$
(U sin g chain rule)
$=-2{\mathrm{sin}}^{2}2x+2{\mathrm{cos}}^{2}2x$
$=-2\left(1-{\mathrm{cos}}^{2}2x\right)+2{\mathrm{cos}}^{2}2x$
$=-2+4{\mathrm{cos}}^{2}2x$
Therefore, the derivative of the given function is
${h}^{\prime }\left(x\right)=-2+4{\mathrm{cos}}^{2}2x$

peterpan7117i

Expert

Step 1
Given function:
$\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)$
$\frac{1}{2}\left(2\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)\right)$
$\frac{1}{2}\mathrm{sin}\left(4x\right)$
Differentiating given function w.r.t x as follows
$\frac{d}{dx}\left(\frac{1}{2}\mathrm{sin}\left(4x\right)\right)$
$=\frac{1}{2}\frac{d}{dx}\left(\mathrm{sin}\left(4x\right)\right)$
$=\frac{1}{2}\mathrm{cos}\left(4x\right)\frac{d}{dx}\left(4x\right)$
$=\frac{1}{2}\mathrm{cos}\left(4x\right)\left(4\right)$
$=2\mathrm{cos}\left(4x\right)$

karton

Expert

Expalnation:
differentiate using the $\frac{\text{product}}{\text{chain rule}}$
$\begin{array}{}\text{Given}\\ y=f\left(x\right)g\left(x\right)\\ \frac{dy}{dx}=f\left(x\right){g}^{\prime }\left(x\right)+g\left(x\right){f}^{\prime }\left(x\right)⇒\text{product rule}\\ f\left(x\right)=\mathrm{sin}2x⇒{f}^{\prime }\left(x\right)=2\mathrm{cos}2x\\ g\left(x\right)=\mathrm{cos}2x⇒{g}^{\prime }\left(x\right)=-2\mathrm{sin}2x\\ \frac{d}{dx}\left(\mathrm{sin}2x\mathrm{cos}2x\right)\\ =-2{\mathrm{sin}}^{2}2x+2{\mathrm{cos}}^{2}2x\\ =2\left({\mathrm{cos}}^{2}2x-{\mathrm{sin}}^{2}2x\right)=2\mathrm{cos}4x\end{array}$

Do you have a similar question?