proof of log derivative using implicit differentiation

Question

How would you use implicit differentiation and the fact that $\log_b x$ is the inverse of $b^x$ to prove that $$\frac {d}{dx} (\log_b x) =\frac{1}{(\ln b)x}$$

Answer 1

Normally we would simply observe that since $\ln(b)$ is a constant:

$$\log_b (x)= \frac{\ln(x)}{\ln(b)} $$ $$ \frac{d \log_b (x)}{dx}= \frac{1}{ \ln(b) x} $$ Since we know how to derive $\ln(x)$, its derivative is $\frac{1}{x}$. But now we wish to use that the logarithm is the inverse of the exponent, then we need to use the rules for exponents. We know how to derive $e^x$ this will be a key property of our method.

Call $u=\log_b (x)$ then since $e^{\ln(b)}=b$ we can write: $$ x= b^u=e^{\ln(b)u} $$ We derive both sides with respect to $x$ using implicit differentiation and the chain rule ($\frac{d}{dx}e^{ax}= a e^{ax}$) remembering that $u$ depends on $x$.: $$ 1= e^{\ln(b) u} \cdot (\ln(b) \frac{d u}{d x} )$$ So now we separate $\frac{d u}{d x}$ from the other terms and get:

$$\frac{d u}{ d x}= \frac{1}{\ln (b) e^{\ln(b) u}}= \frac{1}{\ln(b) x} $$

Answer 2

Hint: set $y=\log_b (x)$ so $$x=b^y$$ Apply natural logarithm to both sides: $$\ln(x)=\ln\left(b^y\right)$$ $$\ln(x)=y\ln(b)$$ From here, use implicit differentiation (Note: $\ln(b)$ is a constant)