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11.(iv)     Find the derivative of the following functions:  \csc x

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Given :

f(x)=\csc x=\frac{1}{\sin x}

Now As we know the quotient rule of derivative,

\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}

So applying this rule, we get

\frac{d(\frac{1}{\sin x})}{dx}=\frac{(\sin x)\frac{d(1)}{dx}-1\frac{d(\sin x)}{dx}}{(\sin x)^2}

\frac{d(\frac{1}{\sin x})}{dx}=\frac{-1(\cos x)}{(\sin x)^2}

\frac{d(\frac{1}{\sin x})}{dx}=-\frac{(\cos x)}{(\sin x)}\frac{1}{\sin x}

\frac{d(\csc x)}{dx}=-\cot x \csc x

Posted by

Pankaj Sanodiya

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