If $$f(x)=g(x)+g(-x)2+2h(x)+h(-x)-1$$,where g and h are differentiable functions then ⨍ '$$(0)=$$

The given function is f $$(x)=g(x)+g(-x)2+2h(x)+h(-x)-1$$,This can be written as $$fx=gx+g-x2+2hx+h-xDifferentiate$$ both sidesf'$$x=g$$'$$x-g$$'$$-x2+2h$$'$$x-h$$'$$-xSubstitute$$ $$x=0f$$'$$0=0$$

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If $f(x)= \frac{g (x) + g (- x)}{2} + \frac{2}{{[h (x) + h (- x)]}^{- 1}},$ where g and h are differentiable functions then $⨍ '(0)=$

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The given function is $f (x)= \frac{g (x) + g (- x)}{2} + \frac{2}{{[h (x) + h (- x)]}^{- 1}},$

This can be written as

$f (x) = \frac{g (x) + g (- x)}{2} + 2 [h (x) + h (- x)]$

Differentiate both sides

$f' (x) = \frac{g' (x) - g' (- x)}{2} + 2 [h' (x) - h' (- x)]$

Substitute $x = 0$

$f' (0) = 0$

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If f(x)=g(x)+g(-x)2+2h(x)+h(-x)-1,where g and h are differentiable functions then ⨍ '(0)=

Unlock the full solution & master the concept.

Ready to Test Your Skills?

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If $f(x)= \frac{g (x) + g (- x)}{2} + \frac{2}{{[h (x) + h (- x)]}^{- 1}},$ where g and h are differentiable functions then $⨍ '(0)=$