smallcrystalslpxs

2023-03-26

How to differentiate $1+{\mathrm{cos}}^{2}\left(x\right)$?

tsunamiwynd6wmb

Beginner2023-03-27Added 7 answers

-sin2x

Explanation:

differentiate using the chain rule

$\text{given}\phantom{\rule{1ex}{0ex}}y=f\left(g\left(x\right)\right)\phantom{\rule{1ex}{0ex}}\text{then}$

$\frac{dy}{dx}=f\prime \left(g\left(x\right)\right)\times g\prime \left(x\right)\leftarrow {\text{chain rule}}$

$y=1+{\left(\mathrm{cos}x\right)}^{2}$

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

${\frac{dy}{dx}}=-2\mathrm{sin}x\mathrm{cos}x=-\mathrm{sin}2x$

Explanation:

differentiate using the chain rule

$\text{given}\phantom{\rule{1ex}{0ex}}y=f\left(g\left(x\right)\right)\phantom{\rule{1ex}{0ex}}\text{then}$

$\frac{dy}{dx}=f\prime \left(g\left(x\right)\right)\times g\prime \left(x\right)\leftarrow {\text{chain rule}}$

$y=1+{\left(\mathrm{cos}x\right)}^{2}$

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

${\frac{dy}{dx}}=-2\mathrm{sin}x\mathrm{cos}x=-\mathrm{sin}2x$

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