 # How do you find the derivative of \cot x ? tebollahb 2021-12-18 Answered
How do you find the derivative of $\mathrm{cot}x$ ?
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Explanation:
$y=\mathrm{cot}x$
$y=\frac{1}{\mathrm{tan}x}$
$y=\frac{1}{\frac{\mathrm{sin}x}{\mathrm{cos}x}}$
$y=\frac{\mathrm{cos}x}{\mathrm{sin}x}$
Letting $y=\frac{g\left(x\right)}{h\left(x\right)}$, we have that $g\left(x\right)=\mathrm{cos}x$ and $h\left(x\right)=\mathrm{sin}x$
${y}^{\prime }=\frac{{g}^{\prime }\left(x\right)×h\left(x\right)-g\left(x\right)×{h}^{\prime }\left(x\right)}{{\left(h\left(x\right)\right)}^{2}}$
${y}^{\prime }=\frac{-\mathrm{sin}x×\mathrm{sin}x-\left(\mathrm{cos}x×\mathrm{cos}x\right)}{{\left(\mathrm{sin}x\right)}^{2}}$
${y}^{\prime }=\frac{-{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}$
${y}^{\prime }=\frac{-\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)}{{\mathrm{sin}}^{2}x}$
${y}^{\prime }=-\frac{1}{{\mathrm{sin}}^{2}x}$
${y}^{\prime }=-{\mathrm{csc}}^{2}x$
Hopefully this helps!

We have step-by-step solutions for your answer! kalfswors0m

As per trigonometric identities, cot x can also be written as $\frac{\mathrm{cos}x}{\mathrm{sin}x}$
Now we can use quotient rule of differentiation to find the derivative of $\mathrm{cot}x$.
$\frac{d\mathrm{cot}x}{dx}=\frac{d\left(\frac{\mathrm{cos}x}{\mathrm{sin}x}\right)}{dx}$
Quotient rule: $\frac{d\left(uv\right)}{dx}=\frac{\left(\frac{vdu}{dx}-\frac{udv}{dx}\right)}{{v}^{2}}$
Here
We can substitute the formulae for the derivatives of sin x and cos x given by
$\frac{d\left(\mathrm{cos}x\right)}{dx}=-\mathrm{sin}x$ and $\frac{d\left(\mathrm{sin}x\right)}{dx}=\mathrm{cos}x$
On putting the above values in equation (ii), we get
$\frac{d\left(\mathrm{cot}x\right)}{dx}=\frac{-\mathrm{sin}x\mathrm{sin}x-\mathrm{cos}x\mathrm{cos}x}{{\mathrm{sin}}^{2}x}$
$=\frac{-\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)}{{\mathrm{sin}}^{2}x}$
$=\frac{-1}{{\mathrm{sin}}^{2}x}$
$=-\left({\mathrm{csc}}^{2}x\right)$
$⇒\frac{d\left(\mathrm{cot}x\right)}{dx}=-{\mathrm{csc}}^{2}x$

We have step-by-step solutions for your answer!