Question

Y = -cot^2 X/2 - 2log sin X/2 prove that dy/dx = cot ^3 x/2

Answer 1

$\textbf{Given:}$

$\mathsf{y=-cot^2\frac{x}{2}-2\;log\,sin\frac{x}{2}}$

$\textbf{To find:}$

$\mathsf{\dfrac{dy}{dx}}$

$\textbf{Solution:}$

$\textsf{We apply chain rule to find the derivative}$

$\mathsf{Consider,}$

$\mathsf{y=-cot^2\frac{x}{2}-2\;log\,sin\frac{x}{2}}$

$\textsf{Differentiate with respect to 'x'}$

$\mathsf{\dfrac{dy}{dx}=-2\,cot\frac{x}{2}\,\left(-cosec^2\frac{x}{2}\right)\frac{1}{2}-2\;\left(\dfrac{1}{sin\frac{x}{2}}\right)cos\frac{x}{2}\;\frac{1}{2}}$

$\mathsf{\dfrac{dy}{dx}=cot\frac{x}{2}\,\left(cosec^2\frac{x}{2}\right)-\left(\dfrac{1}{sin\frac{x}{2}}\right)cos\frac{x}{2}}$

$\mathsf{\dfrac{dy}{dx}=cot\frac{x}{2}\,\left(cosec^2\frac{x}{2}\right)-\left(\dfrac{cos\frac{x}{2}}{sin\frac{x}{2}}\right)}$

$\mathsf{\dfrac{dy}{dx}=cot\frac{x}{2}\,\left(cosec^2\frac{x}{2}\right)-cot\frac{x}{2}}$

$\mathsf{\dfrac{dy}{dx}=cot\frac{x}{2}(cosec^2\frac{x}{2}-1)}$

$\mathsf{\dfrac{dy}{dx}=cot\frac{x}{2}(cot^2\frac{x}{2})}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=cot^3\frac{x}{2}}}$

$\textbf{Find more:}$

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