witglasjanf

2023-02-26

What is the derivative of ${\left(\mathrm{tan}\left(8x\right)\right)}^{2}$?

Dakota George

Answer is $=16\mathrm{tan}\left(8x\right){\mathrm{sec}}^{2}\left(8x\right)$
Solution: We need
$\left({u}^{n}\right)\prime =n{u}^{n-1}\cdot u\prime$
$\left(\mathrm{tan}x\right)\prime ={\mathrm{sec}}^{2}x$
The chain rule is used to compute this derivative.
Let $y={\left(\mathrm{tan}\left(8x\right)\right)}^{2}$
$\frac{dy}{dx}=2\mathrm{tan}\left(8x\right)\cdot {\mathrm{sec}}^{2}\left(8x\right)\cdot 8$
Hence, $=16\mathrm{tan}\left(8x\right){\mathrm{sec}}^{2}\left(8x\right)$

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