Question

Use the chain rule to prove the following.

(a). The derivative of an even function is an odd function.

(b). The derivative of an odd function is an even function.

Short answer

(a). It is proved that the derivative of an even function is an odd function.

(b).It is proved that the derivative of an odd function is an even function.

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

(a) Step 2: The derivative of an even function

Let \(f\left( x \right)\) be an even function then \(f\left( x \right) = f\left( { - x} \right)\). Differentiate both the sides with respect to \(x\) as follows:

\(\begin{array}{l}f'\left( x \right) = f'\left( { - x} \right)\frac{d}{{dx}}\left( { - x} \right)\\f'\left( x \right) = f'\left( { - x} \right)\left( { - 1} \right)\\f'\left( x \right) = - f'\left( x \right)\end{array}\)

The above calculation shows that the derivative of an even function is an odd function.

(b) Step 3: The derivative of an odd function 

Let \(f\left( x \right)\) be an odd function then \(f\left( x \right) = - f\left( { - x} \right)\). Differentiate both the sides with respect to \(x\) as follows:

\(\begin{array}{l}f'\left( x \right) = - f'\left( { - x} \right)\frac{d}{{dx}}\left( { - x} \right)\\f'\left( x \right) = - f'\left( { - x} \right)\left( { - 1} \right)\\f'\left( x \right) = f'\left( { - x} \right)\end{array}\)

The above calculation shows the derivative of an odd function is an even function.

Hence, it is proved.