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