Compute the derivative of g(x) = cot x. | vedclass

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(A) By definition,$g(x) = \cot x = \frac{\cos x}{\sin x}$. We use the quotient rule on this function wherever it is defined.$\frac{dg}{dx} = \frac{d}{

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$-\csc^2 x$

Vedclass · Answer

(A) By definition,$g(x) = \cot x = \frac{\cos x}{\sin x}$. We use the quotient rule on this function wherever it is defined.$\frac{dg}{dx} = \frac{d}{dx}(\cot x) = \frac{d}{dx}\left(\frac{\cos x}{\sin x}\right)$$= \frac{(\cos x)'(\sin x) - (\cos x)(\sin x)'}{(\sin x)^2}$$= \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2}$$= -\frac{\sin^2 x + \cos^2 x}{\sin^2 x} = -\csc^2 x$

Compute the derivative of $g(x) = \cot x$.

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