Shania Houston

2023-03-05

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

Hayden Dudley

The derivative for this is found by the chain rule.
Let, $y=f\left(x\right)={x}^{2\mathrm{cos}x}$.
$\therefore \frac{dy}{dx}=\frac{d}{dx}\left({x}^{2\mathrm{cos}x}\right)\cdot \frac{d}{dx}\left(2\mathrm{cos}x\right)$.
$\therefore \frac{dy}{dx}=\left(2\mathrm{cos}x\cdot {x}^{2\mathrm{cos}x-1}\right)\cdot \left(-2\mathrm{sin}x\right)$.
Hence, $\therefore \frac{dy}{dx}=-4\mathrm{sin}x\mathrm{cos}x\cdot {x}^{2\mathrm{cos}x-1}$.

criticoneitor5nq

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

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