Shania Houston

2023-03-05

How to differentiate $y={x}^{2\mathrm{cos}x}$?

Hayden Dudley

Beginner2023-03-06Added 6 answers

The derivative for this is found by the chain rule.

Let, $y=f\left(x\right)={x}^{2\mathrm{cos}x}$.

$\therefore \frac{dy}{dx}=\frac{d}{dx}\left({x}^{2\mathrm{cos}x}\right)\cdot \frac{d}{dx}\left(2\mathrm{cos}x\right)$.

$\therefore \frac{dy}{dx}=(2\mathrm{cos}x\cdot {x}^{2\mathrm{cos}x-1})\cdot (-2\mathrm{sin}x)$.

Hence, $\therefore \frac{dy}{dx}=-4\mathrm{sin}x\mathrm{cos}x\cdot {x}^{2\mathrm{cos}x-1}$.

Let, $y=f\left(x\right)={x}^{2\mathrm{cos}x}$.

$\therefore \frac{dy}{dx}=\frac{d}{dx}\left({x}^{2\mathrm{cos}x}\right)\cdot \frac{d}{dx}\left(2\mathrm{cos}x\right)$.

$\therefore \frac{dy}{dx}=(2\mathrm{cos}x\cdot {x}^{2\mathrm{cos}x-1})\cdot (-2\mathrm{sin}x)$.

Hence, $\therefore \frac{dy}{dx}=-4\mathrm{sin}x\mathrm{cos}x\cdot {x}^{2\mathrm{cos}x-1}$.

criticoneitor5nq

Beginner2023-03-07Added 3 answers

Use some version of logarithmic differentiation.

Rewriting logarithmic difference as e to a power is my current preferred form.

$y={x}^{2\mathrm{cos}x}={e}^{2\mathrm{cos}x\mathrm{ln}x}$

Thus, use $\frac{d}{dx}\left({e}^{u}\right)={e}^{u}\frac{du}{dx}$ to get

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

$={x}^{2\mathrm{cos}x}\cdot [\frac{2\mathrm{cos}x}{x}-2\mathrm{sin}x\mathrm{ln}x]$

Rewriting logarithmic difference as e to a power is my current preferred form.

$y={x}^{2\mathrm{cos}x}={e}^{2\mathrm{cos}x\mathrm{ln}x}$

Thus, use $\frac{d}{dx}\left({e}^{u}\right)={e}^{u}\frac{du}{dx}$ to get

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

$={x}^{2\mathrm{cos}x}\cdot [\frac{2\mathrm{cos}x}{x}-2\mathrm{sin}x\mathrm{ln}x]$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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