Question

find dy/dx y=cot^3[log(x^3)]​

Answer 1

$\textbf{Given:}$

$y=cot^3[log(x^3)]$

$\textbf{To find:}$

$\dfrac{dy}{dx}$

$\textbf{Solution:}$

$\text{I have applied chain rule to differentiate the given function}$

$y=cot^3[log(x^3)]$

$\text{Differentiate with respect to x}$

$\dfrac{dy}{dx}=3\,cot^2[log(x^3)](-cosec^2[log(x^3)])(\dfrac{1}{x^3})(3x^2)$

$\dfrac{dy}{dx}=3\,cot^2[log(x^3)](-cosec^2[log(x^3)])(\dfrac{3}{x})$

$\dfrac{dy}{dx}=\dfrac{-9\,cot^2[log(x^3)](cosec^2[log(x^3)])}{x}$

$\text{Using}$

$\boxed{cosec^2A=1+cot^2A}\;\;\text{we get\,}$

$\dfrac{dy}{dx}=\dfrac{-9\,cot^2[log(x^3)](1+cot^2[log(x^3)])}{x}$

$\dfrac{dy}{dx}=\dfrac{-9}{x}[cot^2[log(x^3)]+cot^4[log(x^3)]]$

$\therefore\boxed{\bf\dfrac{dy}{dx}=\dfrac{-9}{x}[cot^2[log(x^3)]+cot^4[log(x^3)]]}$

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