Question

Prove that derivative of even function is odd function and derivative of odd function is even function​

Answer 1

$\large\underline{\sf{Solution-a}}$

To prove that derivative of even function is odd.

Let assume that f(x) be even function.

We know,

A function f(x) is said to be even function iff f(- x) = f(x).

So, we have

$\rm \: f( - x) = f(x)$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx} f( - x) =\dfrac{d}{dx} f(x)$

$\rm \: f'( - x)\dfrac{d}{dx}( - x) = f'(x)$

$\rm \: f'( - x)( - 1) = f'(x)$

$\rm \: - f'( - x) = f'(x)$

$\rm \: f'( - x) \: = \: - \: f'(x)$

We know

A function f(x) is said to be odd function iff f(- x) = - f(x)

So, using this definition,

$\rm\implies \:f'(x) \: is \: an \: odd \: function$

$\large\underline{\sf{Solution-b}}$

To prove that derivative of odd function is even.

Let assume that f(x) be an odd function.

So, by definition of odd function.

$\rm \: f( - x) = - f(x)$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx} f( - x) \: = \: - \: \dfrac{d}{dx} f(x)$

$\rm \: f'( - x)( - 1) \: = \: - \: f'(x)$

$\rm \: f'( - x) \: = \: f'(x)$

So, by definition of even function, we get

$\rm\implies \:f'(x) \: is \: an \: even \: function$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\sf \underline \purple{Solution:−}$

$: \longmapsto\rm{f \: is \: an \: even \: function. }$

$\longmapsto\rm{f ( - x) = f(x)}$

Take the derivative both sides :-

$\longmapsto\rm{ \frac{d}{dx}f( - x) = \frac{d}{dx} f(x) }$

Use chain rule :-

$\longmapsto\rm{ \frac{d}{dx} f( - x) \frac{d}{dx} ( - x) = f'(x) }$

$\longmapsto\rm{f'( - x)( - 1) = f'(x)}$

$\longmapsto\rm{f'( - x) = - f'(x)}$

$:\longmapsto\rm{f' \: is \: a \: odd \: function. }$

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ADDITIONAL INFORMATION :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}\end{gathered}$

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@Shivam

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