Susan Munoz

2021-11-16

Find the derivative of this problem.
$h\left(x\right)=\mathrm{sin}2x\mathrm{cos}2x$

Thomas Conway

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

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