nemired9

Answered

2021-12-26

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

Answer & Explanation

Orlando Paz

Expert

2021-12-27Added 42 answers

Step 1

Product Rule of differentiation:

If f(x) and g(x) are any two differentiable functions, then

${f\left(g\right)}^{\prime}\left(x\right)=f\left(x\right)\times {g}^{\prime}\left(x\right)+{f}^{\prime}\left(x\right)\times g\left(x\right)$

Chain rule of differentiation:

$\frac{d}{dx}f\left(g\left(x\right)\right)={f}^{\prime}\left(g\left(x\right)\right)\times \frac{d}{dx}g\left(x\right)$

$={f}^{\prime}\left(g\left(x\right)\right)\times {g}^{\prime}\left(x\right)\times \frac{d}{dx}\left(x\right)$

$={f}^{\prime}\left(g\left(x\right)\right)\times {g}^{\prime}\left(x\right)$

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)

$=\mathrm{sin}2x(-\mathrm{sin}2x)\left(2\right)+\mathrm{cos}2x\left(\mathrm{cos}2x\right)\left(2\right)$

(U sin g chain rule)

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

$=-2(1-{\mathrm{cos}}^{2}2x)+2{\mathrm{cos}}^{2}2x$

$=-2+4{\mathrm{cos}}^{2}2x$

Therefore, the derivative of the given function is

${h}^{\prime}\left(x\right)=-2+4{\mathrm{cos}}^{2}2x$

Product Rule of differentiation:

If f(x) and g(x) are any two differentiable functions, then

Chain rule of differentiation:

The given function is

Step 2

Differentiate the given function using the product rule as follows.

(U sin g product rule)

(U sin g chain rule)

Therefore, the derivative of the given function is

peterpan7117i

Expert

2021-12-28Added 39 answers

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

$\frac{d}{dx}\left(\frac{1}{2}\mathrm{sin}\left(4x\right)\right)$

$=\frac{1}{2}\frac{d}{dx}\left(\mathrm{sin}\left(4x\right)\right)$

$=\frac{1}{2}\mathrm{cos}\left(4x\right)\frac{d}{dx}\left(4x\right)$

$=\frac{1}{2}\mathrm{cos}\left(4x\right)\left(4\right)$

$=2\mathrm{cos}\left(4x\right)$

Given function:

Differentiating given function w.r.t x as follows

karton

Expert

2022-01-04Added 439 answers

Expalnation:

differentiate using the

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