#### To determine

**To show:** The derivative of an even function is an odd function.

#### Explanation

**Definition used:**

(i) If f(x) even function, then f(−x)=f(x).

(ii) If f(x) odd function, then f(−x)=−f(x).

**Result used:** *Chain Rule*

If *h* is differentiable at *t* and *g* is differentiable at h(t), then the composite function F=g∘h defined by F(t)=g(h(t)) is differentiable at *t* and F′ is given by the product

F′(t)=g′(h(t))⋅h′(t) (1)

**Proof:**

Let f(x) is even function.

Then, f(−x)=f(x).

The derivative of f(x) is computed as follows,

f′(x)=ddx(f(x))=ddx(f(−x)) (Q f(x)=f(−x))

Let h(x)=−x and g(u)=f(u) where u=h(x).

Apply the chain rule as shown in equation (1),

f′(x)=g′(h(x))⋅h′(x) (2)

The derivative g′(h(x)) is computed as follows,

g′(h(x))=g′(u)=ddu(g(u))=ddu(f(u))=f′(u)

Substitute u=−x in the above equation,

g′(h(x))=f′(−x)

Thus the derivative g′(h(x))=f′(−x).

The derivative of h(x) is computed as follows,

h′(x)=ddx(−x)=(−1x1−1)=−1

Thus, the derivative of h(x) is h′(x)=−1.

Substitute f′(−x) for g′(h(x)) and −1 for h′(x) in equation (2),

f′(x)=f′(−x)(−1)=−f′(−x)

Hence, the derivative of even function is odd function is proved.

#### To determine

**To show:** The derivative of an odd function is an even function.

#### Explanation

**Proof:**

Let f(x) is odd function.

Then, f(−x)=−f(x).

Differentiate with respect to *x*,

ddx[f(−x)]=ddx[−f(x)]

The derivative of −f(x) is computed as follows,

ddx[−f(x)]=−ddx[f(x)]

=−f′(x) (3)

The derivative of f(−x) is computed as follows,

Let h(x)=−x and g(u)=f(u) where u=h(x).

Apply the chain rule as shown in equation (1),

ddx[f(−x)]=g′(h(x))⋅h′(x) (4)

The derivative g′(h(x)) is computed as follows,

g′(h(x))=g′(u)=ddu(g(u))=ddu(f(u))=f′(u)

Substitute u=−x in the above equation,

g′(h(x))=f′(−x)

Thus the derivative g′(h(x))=f′(−x).

The derivative of h(x) is computed as follows,

h′(x)=ddx(−x)=(−1x1−1)=−1

Thus, the derivative of h(x) is h′(x)=−1.

Substitute f′(−x) for g′(h(x)) and −1 for h′(x) in equation (4),

ddx[f(−x)]=f′(−x)(−1)

=−f′(−x) (5)

From (3) and (5), −f′(x)=−f′(−x).

Multiply by −1, f′(x)=f′(−x).

Hence, the derivative of odd function is even function is proved.