Karl's Calculus Tutor - Solution to Exercise 6.2-6

Box 6.2-6: Solution to Exercise 6.2-6KCT logo

© 1997 by Karl Hahn

The problem was to use implicit differentiation to determine the derivative of f(x) when you know that 

   bf(x)  =  xn
To do implicit differentiation, we take the derivative of both sides, and to take the derivative of the left side, we see that it is a composite of bx and f(x). If you let g(x) = bx, then the left side is g(f(x)). This tells us to use the chain rule. To apply the chain rule, you first have to know that (in accordance with the method you learned in this section) 
   g'(x)  =  ln(b) bx
We do not yet have an expression for f'(x), so we just write f'(x) wherever it occurs.

The chain rule says that the derivative of the left side is 

   g'(f(x)) f'(x)
Substituting the expression we have for g'(x), this becomes 
   ln(b) bf(x) f'(x)
And that's what we get on the left side. On the right we have to take the derivative of xn. You learned how to do that in the early sections on derivatives, and you know that its derivative is nxn-1. So, putting the right and left sides together, we have 
   ln(b) bf(x) f'(x)  =  nxn-1
But the original equation told us that 
   bf(x)  =  xn
So we can substitute xn wherever we see bf(x). This gives 
   ln(b) xn f'(x)  =  nxn-1
Now simply solve for f'(x). Divide out the ln(b) and the xn, and f'(x) will be left standing alone on one side of the equation. 
              nxn-1
   f'(x)  =          
             ln(b) xn
Look at the powers of x in the numerator and denominator. We have one more x in the denominator than in the numerator. So we can cancel all the x's in the numerator with all but one of them in the denominator, leaving 
                n
   f'(x)  =         
             ln(b) x
We started with an exponential expression and we are left with this simple expression as the derivative. It's nothing more than a constant times 1/x. This is the first example we have had that shows how calculus connects so called transcendental functions (which include logs, exponentials, trig functions, and many more) to ordinary algebraic functions such as 1/x.

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