Question

Differentiate cos{log(cot)}^7

with respect to X.​

Answer 1

$\textbf{Given:}$

$\mathsf{cos(log(cotx))^7}$

$\textbf{To find:}$

$\mathsf{Derivative\;of\;cos(log(cotx))^7}$

$\textbf{Solution:}$

$\textsf{We apply chain rule diffentiate the given function}$

$\mathsf{Let\;y=cos(log(cotx))^7}$

$\textsf{Differentiate with respect to x}$

$\mathsf{\dfrac{dy}{dx}=-sin(log(cotx))^7\;7(log(cotx))^6\;\dfrac{1}{cotx}(-cosec^2x)}$

$\mathsf{\dfrac{dy}{dx}=-sin(log(cotx))^7\;7(log(cotx))^6\;tanx(-cosec^2x)}$

$\mathsf{\dfrac{dy}{dx}=-sin(log(cotx))^7\;7(log(cotx))^6\;\dfrac{sinx}{cosx}\left(\dfrac{-1}{sin^2x}\right)}$

$\mathsf{\dfrac{dy}{dx}=-sin(log(cotx))^7\;7(log(cotx))^6\;\left(\dfrac{-1}{cosx\;sinx}\right)}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{sin(log(cotx))^7\;7(log(cotx))^6}{cosx\;sinx}}}$

$\textbf{Find more:}$

1.y = (tan x +cot x)/ (tan x - cotx ) then dy/dx =?

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3.If y = tan^–1(sec x + tan x), then find dy/dx.

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4.If y= log tan(π/4+x/2),show that dy/dx=secx

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5.Differentiate w. r. t.x.

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Answer 2

$\textbf{Given:}$

$\mathsf{cos(log(cotx))^7}$

$\textbf{To find:}$

$\mathsf{Derivative\;of\;cos(log(cotx))^7}$

$\textbf\red{Solution:}$

$\textsf{We apply chain rule diffentiate the given function}$

$\mathsf{Let\;y=cos(log(cotx))^7}$

$\textsf{Differentiate with respect to x}$

$\mathsf{\dfrac{dy}{dx}=-sin(log(cotx))^7\;7(log(cotx))^6\;\dfrac{1}{cotx}(-cosec^2x)}$

$\mathsf{\dfrac{dy}{dx}=-sin(log(cotx))^7\;7(log(cotx))^6\;tanx(-cosec^2x)}$

$\mathsf{\dfrac{dy}{dx}=-sin(log(cotx))^7\;7(log(cotx))^6\;\dfrac{sinx}{cosx}\left(\dfrac{-1}{sin^2x}\right)}$

$\mathsf{\dfrac{dy}{dx}=-sin(log(cotx))^7\;7(log(cotx))^6\;\left(\dfrac{-1}{cosx\;sinx}\right)}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{sin(log(cotx))^7\;7(log(cotx))^6}{cosx\;sinx}}}$