Vtyjkyuk

I understand

$'ln(x) = frac1'e^ln(x) =frac1x $

But if I use the chain rule in the denominator to calculate the derivative

$'e^ln(x) => ('e^ln(x))('ln(x))=(x)(frac1x)=1$

But then

$frac1xneqfrac11$

What am I doing wrong?

The devil is in the detail. In your first equation it must be 1/e^Ln(x)=1/x. For the following reason: suppose f,g are differentiable. The chain rule then says:

f(g(x))’=f’(g(x))*g’(x)

So if g is the inverse function of f, f(g(x))=x then differentiation gives:

1= f’(g(x))*g’(x) I.e.

g’(x)=1/f’(g(x)) when the denominator is nonzero.

The key point here is that you take the derivative of the “outer” function and leave the “internal” function. Applying this to your case gives the result. That is f(y)=e^y and g(x)=ln(x)

$f(x)=e^x$ and $f^-1(x)=ln(x)$. By direct computation we see that

$(f^-1)'(x)=d/dx(ln(x))=1/x$

By the inverse function rule, this should be equal to $frac1f'(f^-1(x))$. Well, $f'(x)=e^x$ and hence $frac1f'(f^-1(x))=frac1e^ln(x)=frac1x$ as desired.