If $f(x) = \frac{g(x) + g( - x)}{2}$+$\frac{2}{\left\lbrack h(x) + h( - x) \right\rbrack^{- 1}}\$where g and $h$ are differentiable function, then $f^{'}(0)$ is

Question

Dr. Abubacker · Accepted Answer

(d)
Given, $f(x) = \frac{g(x) + g( - x)}{2} + \frac{2}{\left\lbrack h(x) + h( - x) ightbrack^{- 1}}$
$\Rightarrow (x) = \frac{g(x) + g( - x)}{2} + 2\left\lbrack h(x) + h( - x) ightbrack$
On differentiating w.r.t. $x,\$we get
$f^{'}(x) = \frac{g^{'}(x) - g^{'}( - x)}{2} + 2\lbrack h^{'}(x) - h^{'}( - x)brack$
$	herefore{\ f}^{'}(0) = \frac{g^{'}(0) - g^{'}(0)}{2} + 2\lbrack h^{'}(0) - h^{'}(0)brack$
= 0

Q. If $f(x) = \frac{g(x) + g( - x)}{2}$+$\frac{2}{\left\lbrack h(x) + h( - x) \right\rbrack^{- 1}}\$where g and $h$ are differentiable function, then $f^{'}(0)$ is

Explanation

Reviewed by Dr. Abubacker

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