 efedrinalvog

2023-02-26

How to find the derivative of $y=\mathrm{tan}x+\mathrm{cot}x$? Kylie Woodward

Explanation: Separately address the tanx and the cotx. Don't forget about trig function derivatives:
$\frac{d}{dx}\left(\mathrm{sin}\left(x\right)\right)=\mathrm{cos}\left(x\right)\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{tan}\left(x\right)\right)={\mathrm{sec}}^{2}\left(x\right)\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{sec}\left(x\right)\right)=\mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{cos}\left(x\right)\right)=-\mathrm{sin}\left(x\right)\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{cot}\left(x\right)\right)=-{\mathrm{csc}}^{2}\left(x\right)\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\mathrm{csc}\left(x\right)\right)=-\mathrm{csc}\left(x\right)\mathrm{cot}\left(x\right)$
Derivative of tan x is ${\mathrm{sec}}^{2}x$. Derivative of cot x is $-{\mathrm{csc}}^{2}x$. Since they are adding together, we can treat them it as tan'x and cot'(x):
$f\prime \left(x\right)={\mathrm{sec}}^{2}\left(x\right)-{\mathrm{csc}}^{2}\left(x\right)$ tammypierce5kgk

Solution:
A slightly different approach...
Rewrite in terms of sine and cosine.
$y=\frac{\mathrm{sin}x}{\mathrm{cos}x}+\frac{\mathrm{cos}x}{\mathrm{sin}x}$
$y=\frac{{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x}{\mathrm{sin}x\mathrm{cos}x}$
$y=\frac{1}{\mathrm{sin}x\mathrm{cos}x}$
$y={\left(\mathrm{sin}x\mathrm{cos}x\right)}^{-1}$
Step 2
To differentiate, apply the chain rule. Let $y={u}^{-1}$ and $u=\mathrm{sin}x\mathrm{cos}x$. The product rule can be used to differentiate the function u.
$\frac{du}{dx}=\mathrm{cos}x\left(\mathrm{cos}x\right)+\mathrm{sin}x\left(-\mathrm{sin}x\right)={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=\mathrm{cos}2x$ By the power rule, $\frac{dy}{du}=-\frac{1}{{u}^{2}}$
$y\prime =\frac{dy}{du}\cdot \frac{du}{dx}$
$y\prime =\mathrm{cos}2x\cdot -\frac{1}{{u}^{2}}$
$y\prime =-\frac{\mathrm{cos}\left(2x\right)}{{\left(\mathrm{sin}x\mathrm{cos}x\right)}^{2}}$
Answer is $y\prime =-\mathrm{cos}2x{\mathrm{sec}}^{2}x{\mathrm{csc}}^{2}x$

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