In an earlier lecture, we did an abbreviated proof of the Power Rule, $D_x(x^n) = nx^{n-1}$, where $n$ was an integer. We can use implicit differentiation to show that this is true for all rational numbers. (We'll show the general case for all real numbers in the next lecture.) Start with $y=x^{p/q}$ where $p$ and $q$ are integers and see if you can work through the process to show that $y'=\frac{p}{q}x^{p/q-1}$. If you need help or suggestions, you can click on the boxes below to see the individual steps.
Start with $y=x^{p/q}$ and eliminate the fraction in the exponent by raising both sides to an appropriate power.
$$y^q=\left(x^{p/q}\right)^q$$ $$y^q=x^p$$
Now differentiate both sides.
$$q y^{q-1} y'=px^{p-1}$$
Rewrite the left side of the equation with two y's by splitting up the exponent.
$$q y^{q} y^{-1} y'=px^{p-1}$$
Replace $y^q$ with $x^p$. (They're equal from step 1.)
$$q x^{p} y^{-1} y'=px^{p-1}$$
Solve the equation for y'.
$$y'=\frac{y px^{p-1}}{qx^p}$$
Simplify the p's by subtracting the exponents then replace $y$ with $x^{p/q}$.
