Kylee Mathis

2023-02-26

What is the derivative of $\therefore \frac{dy}{dx}=-\mathrm{cot}x\mathrm{csc}x$?

Gerald Dickerson

Since $f\left(x\right)=\frac{1}{\mathrm{sin}\left(x\right)}=\mathrm{csc}\left(x\right)$, the answer can be written down from memorization that $f\prime \left(x\right)=-\mathrm{csc}\left(x\right)\mathrm{cot}\left(x\right)$.
Alternatively, the Quotient Rule can be used:
$f\prime \left(x\right)=\frac{\mathrm{sin}\left(x\right)\cdot 0-1\cdot \mathrm{cos}\left(x\right)}{{\mathrm{sin}}^{2}\left(x\right)}=-\frac{\mathrm{cos}\left(x\right)}{{\mathrm{sin}}^{2}\left(x\right)}$
$=-\frac{1}{\mathrm{sin}\left(x\right)}\cdot \frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}=-\mathrm{csc}\left(x\right)\mathrm{cot}\left(x\right)$

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