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11(ii)   Find the derivative of the following functions:  \sec x

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best_answer

Given

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

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

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

\frac{d(\sec x)}{dx}=\tan x\sec x

Posted by

Pankaj Sanodiya

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