If fx - x3 - -tanx- then f- -2is equal to- fx - x3 - -tanx- - f- -2 - - -

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Definition of a Determinant

If fx = x3 + ...

Question

If f(x) = x³ + |tanx| then $f^{'} (\frac{π}{2})$ is equal to

यदि f(x) = x³ + |tanx|, तब $f^{'} (\frac{π}{2})$ का मान है

\frac{3 π^{2}}{4}

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Does not exist

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\frac{π^{2}}{4}

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Solution

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Similar questions

y = {(cos x)}^{x^{2}}

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{tan}^{- 1} (x) + {tan}^{- 1} (\sqrt{3}) = \frac{π}{2}

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f (x)

x

g (x) = f (x) - f^{'} (x) + f^{''} (x)

x

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(\frac{6}{5})}^{\frac{t a n 6 x}{t a n 5 x}}; 0 < x < \frac{π}{2} a + 2; x = \frac{π}{2} (1 + | c o t x |)^{\frac{b | t a n x |}{a}}; \frac{π}{2} < x < π \end{matrix}

x = \frac{π}{2}

a

b

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