#### To determine

**To find:** The composite function in the form f(g(x)) and obtain the derivative of *y*.

#### Answer

The inner function is u=πx and the outer function is f(u)=tanu.

The derivative of *y* is y′(x)=πsec2πx_.

#### Explanation

**Given:**

The function is y=tanπx.

**Formula used:**

**The Chain Rule:**

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

F′(x)=g′(h(x))⋅h′(x) (1)

**Power Rule:**

If *n* is positive integer, then ddx(xn)=nxn−1 (2)

**Calculation:**

Let the inner function be u=g(x) and the outer function be y=f(u).

Then, g(x)=πx and f(u)=tanu. That is,

y=tanπx=f(πx)=f(g(x))

Therefore, y=f(g(x)).

Hence, the inner function is u=πx and the outer function is f(u)=tanu.

Thus, the required form of composite function is f(g(x))=tanu.

Obtain the derivative of *y*.

Let h(x)=πx and g(u)=tanu where u=h(x).

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

y′(x)=g′(h(x))⋅h′(x) (3)

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

g′(h(x))=g′(u)=ddu(g(u))=ddu(tanu)=sec2u

Substitute u=πx in the above equation,

g′(h(x))=sec2πx

Thus, the derivative g′(h(x)) is g′(h(x))=sec2πx.

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

h′(x)=ddx(πx)

Apply the power rule as shown in equation (2),

h′(x)=π[1x1−1]=π(1x0)=π(1)=π

Substitute sec2πx for g′(h(x)) and π for h′(x) in equation (3),

g′(h(x))⋅h′(x)=sec2πx(π)=πsec2πx

Therefore, the derivative of y=tanπx is y′(x)=πsec2πx_.