How do you find the derivative of cot⁡x ?

Question

Thomas White · Accepted Answer

Explanation:y=cot⁡xy=1tan⁡xy=1sin⁡xcos⁡xy=cos⁡xsin⁡xLetting y=g(x)h(x), we have that g(x)=cos⁡x and h(x)=sin⁡xy′=g′(x)×h(x)−g(x)×h′(x)(h(x))2y′=−sin⁡x×sin⁡x−(cos⁡x×cos⁡x)(sin⁡x)2y′=−sin2x+cos2xsin2xy′=−(sin2x+cos2x)sin2xy′=−1sin2xy′=−csc2xHopefully this helps!

kalfswors0m · Accepted Answer

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

How do you find the derivative of \cot x ?

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