Question

Differentiate log (cosec x + cot x)

Answer 1

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Step-by-step explanation:

Derivative of cosec x + cot x is=-cosec\,x\:(cot\,x-cosec\,x)=−cosecx(cotx−cosecx)

Step-by-step explanation:

Given function = cosec x + cot x

We have to find derivative if the function.

\frac{\mathrm{d}(cosec\,x+cot\,x)}{\mathrm{d} x}

dx

d(cosecx+cotx)

=\frac{\mathrm{d}(cosec\,x)}{\mathrm{d} x}+\frac{\mathrm{d}(cot\,x)}{\mathrm{d} x}=

dx

d(cosecx)

+

dx

d(cotx)

=-cosec\,x\:cot\,x+(-cosec^2\,x)=−cosecxcotx+(−cosec

2

x)

=-cosec\,x\:cot\,x-cosec^2\,x=−cosecxcotx−cosec

2

x

=-cosec\,x\:(cot\,x-cosec\,x)=−cosecx(cotx−cosecx)

Therefore, Derivative of cosec x + cot x is =-cosec\,x\:(cot\,x-cosec\,x)=−cosecx(cotx−cosecx)

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Answer 2

$let \: \: \: \: \: y = log(cosecx + cotx) \\ \\ \frac{dy}{dx} = \frac{1}{cosecx + cotx} \frac{d}{dx} (cosecx + cotx) \\ \\ \frac{dy}{dx} = \frac{1}{cosecx + cotx} ( - cosecxcotx - {cosec}^{2} x) \\ \\ \frac{dy}{dx} = \frac{ - cosecx(cotx + cosecx)}{cosecx + cotx} \\ \\ \frac{dy}{dx} = - cosecx$