Implicit Diﬀerentiation (Rational Exponent Rule) We know that if n is an integer then the derivative of y = xn is nxn−1 . Does this formula still work if n is not an integer? I.e. is it true that: d a (x ) = axa−1 . dx We proved this formula using the deﬁnition of the derivative and the binomial theorem for a = 1, 2, ... From this, we also got the formula for a = −1, −2, .... Now we’ll extend this formula to cover rational numbers a = m as well. In n √ n x = x1/n particular, this will let us take the derivative of y = . m dy Suppose y = x n , where m and n are integers. We want to compute . dx None of the rules we’ve learned so far seem helpful here, and if we use the deﬁnition of the derivative we’ll get stuck trying to simplify (x − Δx)m/n . We need a new idea. The thing that’s keeping us from using the deﬁnition of the derivative is that the denominator of n in the exponent forces us to take the nth root of x. We could solve this problem by raising both sides of the equation to the nth power: y yn yn yn m = xn m = (x n )n m = x n ·n = xm What happens if we try to take the derivative now by applying the operator We have a rule for ﬁnding the derivative of a variable raised to an integer power; we can use this rule on both sides of the equation y n = xm . d dx ? yn = d n y dx = xm d m x dx d n How do we compute dx y ? We know that y is a function of x, so we can apply the chain rule with outside function y n and inside function y. Suppose u = y n . Then the chain rule tells us: du du dy = dx dy dx So d n y = dx � d n y dy � dy dy = ny n−1 . dx dx On the right hand side of the equation we have up with: d n y dx ny n−1 = d m dx x = mxm−1 , so we end d m x dx dy = mxm−1 dx dy We’re left with only one unknown quantity in this equation — dx — which dy is exactly what we’re trying to ﬁnd. Can we solve for and use this to ﬁnd dx m/n the derivative of y = x ? We can, but we need to use a lot of algebra to do it. By dividing both sides by ny n−1 we get: dy m xm−1 = dx n y n−1 This looks promising but we want our answer in terms of x, without any y’s mixed in. To get rid of the y we can now substitute xm/n for y. (We couldn’t have done this before taking the derivative because we don’t know how to take the derivative of xm/n — that’s the whole point!) dy dx = = = = = = = = = = So, dy dx = � � m xm−1 n y n−1 � � m xm−1 n (xm/n )(n−1) � � m xm−1 n x(m/n)·(n−1) m xm−1 n xm(n−1)/n m ((m−1)− m(n−1) ) n x n m n(m−1) − m(n−1) n x n n m n(m−1)−m(n−1) n x n m nm−n−nm+m n x n m m−n x n n m (m − n) n x n n m ( m − 1) x n n This is the answer we were hoping to get! We now know that for any rational number a, the derivative of xa is axa−1 .