Question

D/DX of root of cot square of x

Answer 1

$\implies \dfrac{d( {cot}^{2}x) }{dx}$

$\implies \dfrac{d( {cot}^{2}x) }{dx} = 2cot \: x \times \dfrac{d( {cot}x) }{dx} \times \dfrac{dx }{dx}$

$\implies \dfrac{d( {cot}^{2}x) }{dx} = 2cot \: x \times ( - { \csc}^{2} x )\times 1$

$\implies \dfrac{d( {cot}^{2}x) }{dx} = - 2cot x \times{\csc}^{2} x$

$\implies \dfrac{d( {cot}^{2}x) }{dx} = - 2 \: cot x \: {\csc}^{2} x$

Answer :

$- 2 \: cot x \: {\csc}^{2} x$

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Learn more :

d(x^n)/dx = n x^(n-1)

d(log x)/dx = 1/x

d(e^x)/dx = e^x

d(sinx)/dx = cosx

d(cos x)/dx = -sin x

d(cosec x)/dx = -cot x cosec x

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x

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