Question

The Power Rule can be proved using implicit differentiatio\(n = \frac{p}{q}\)n for the case where n is a rational number, , and \(y = f\left( x \right) = {x^n}\) is assumed beforehand to be a differential function. If \(y = {x^{\frac{p}{q}}}\), then \({y^q} = {x^q}\). Use implicit differentiation to show that

\(y' = \frac{p}{q}{x^{\left( {\frac{p}{q}} \right) - {\bf{1}}}}\)

Short answer

The value of \(y'\) is \(\frac{p}{q}{x^{\left( {\frac{p}{q}} \right) - 1}}\).

Step-by-step solution

Differentiation the given equation

Differentiate the equation \({y^q} = {x^p}\) with respect to \(x\) on both sides.

\(\begin{array}{c}q{y^{q - 1}}y' = p{x^{p - 1}}\\y' = \frac{{p{x^{p - 1}}}}{{q{y^{q - 1}}}}\end{array}\)

Reduce the expression for \(y'\) in the required form

The equation \(y' = \frac{{p{x^{p - 1}}}}{{q{y^{q - 1}}}}\) can be simplified as follows:

\(\begin{array}{c}y' = \frac{{p{x^{p - 1}}}}{{q{y^{q - 1}}}} \times \frac{y}{y}\\ = \frac{{p{x^{p - 1}}\left( {{x^{\frac{p}{q}}}} \right)}}{{q{y^q}}}\\ = \frac{{p{x^p}{x^{\left( {\frac{p}{q}} \right) - 1}}}}{{q{x^p}}}\\ = \frac{p}{q}{x^{\left( {\frac{p}{q}} \right) - 1}}\end{array}\)

So, the value of \(y'\) is \(\frac{p}{q}{x^{\left( {\frac{p}{q}} \right) - 1}}\).