JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

f\left( x \right) = \sqrt {1 + {{\cos }^2}{{\left( x \right)}^2}}

, then the value of

f'\left( {\frac{{\sqrt \mathrm{\pi } }}{2}} \right)

View answer and explanation

Correct option: B

- \sqrt {\frac{\mathrm{\pi }}{6}}

Option B (

- \sqrt {\frac{\mathrm{\pi }}{6}}

) is correct because it is the value of the derivative of the given function evaluated at

x = \frac{{\sqrt \mathrm{\pi } }}{2}

More Questions from DIFFERENTIATION

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\frac{{dy}}{{dx}}

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v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}