allhvasstH

2021-09-20

Solve the derivatives.

$z={e}^{\mathrm{cos}\left(2x\right)}$

hosentak

Skilled2021-09-21Added 100 answers

Step 1

Given:

$z={e}^{\mathrm{cos}\left(2x\right)}$

Derivation on both sides

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

Applying exponential power rule

${\left[{e}^{u\left(x\right)}\right]}^{\prime}={e}^{u\left(x\right)}\cdot {u}^{\prime}\left(x\right)$

Step 2

$={e}^{\mathrm{cos}\left(2x\right)}\cdot \frac{d}{dx}\left[\mathrm{cos}\left(2x\right)\right]$

$={e}^{\mathrm{cos}\left(2x\right)}(-\mathrm{sin}\left(2x\right))\cdot \frac{d}{dx}\left[2x\right]$

$=-{e}^{\mathrm{cos}\left(2x\right)}\cdot 2\cdot \frac{d}{dx}\left[x\right]\cdot \mathrm{sin}\left(2x\right)$

$=-{e}^{\mathrm{cos}\left(2x\right)}\cdot 2\cdot 1\cdot \mathrm{sin}\left(2x\right)$

Therefore

$\frac{dz}{dx}=-2{e}^{\mathrm{cos}\left(2x\right)}\mathrm{sin}\left(2x\right)$

Given:

Derivation on both sides

Applying exponential power rule

Step 2

Therefore

Jeffrey Jordon

Expert2022-02-01Added 2605 answers

Answer is given below (on video)

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