#### To determine

**To show:** The derivative of the given function is y′(x)=pqxpq−1 by using implicit differentiation.

#### Explanation

The derivative of the function is dydx=pqxpq−1.

**Given:**

The given function is y=xn, where n=pq is a rational number.

**Derivative rule:** *Chain rule*

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydu⋅dudx.

**Proof:**

Rewrite the given function,

y=xpqyq=xp

Differentiate implicitly with respect to *x*,

ddx(yq)=ddx(xp)ddx(yq)=pxp−1

Apply the chain rule and simplify the terms,

ddy(yq)dydx=pxp−1qyq−1dydx=pxp−1dydx=pxp−1qyq−1

Substitute y=xpq,

dydx=pxp−1q(xpq)q−1=pxp−1q(xpq)q−1=pqxp−1xp−pq=pqxp−1x−p+pq

Simplify the terms and obtain the dydx,

dydx=pqxp−1−p+pq=pqxpq−1

Therefore, the derivative of the function is dydx=pqxpq−1.

Hence, the required result is obtained.