derivatives of general exponential and logarithmic functions

Let \[y=\log_{b}(x)\], then \[b^{y}=x\]. It follows that \[\ln(b^{y})=\ln(x)\], then \[y\ln(b)=\ln(x)\], thus \[y=\ln(x)\cdot\frac{1}{\ln(b)}\].

Differentiating with \[y=\ln(x)\cdot \frac{1}{\ln(b)}\] with respect to \[x\], while keeping in mind that \[\ln(b)\] is a constant, we get \[y^{\prime}=\frac{1}{x}\cdot \frac{1}{\ln(b)}=\frac{1}{x\cdot\ln(b)}\], as per the definition stated in the derivative of natural logarithm function.

More generally, let \[y=\log_{b}(f(x))\], where \[f(x)>0\], then \[b^{y}=f(x)\]. Therefore, \[\ln(b^{y})=\ln(f(x))\], \[y=\ln(f(x))\cdot \frac{1}{\ln(b)}\]. Using chain rule, \[y^{\prime}=\frac{1}{f(x)}\cdot f^{\prime}(x)\cdot \frac{1}{\ln(b)}=\frac{f^{\prime}(x)}{f(x)\cdot\ln(b)}\].

Let \[y=b^{x}\], then \[\ln(y)=x\ln(b)\]. Using implicit differentiation while keeping in mind \[\ln(b)\] is a constant, we define \[y=g(x)\], \[(\ln(g(x)))^{\prime}=(x\ln(b))^{\prime}\], thus becoming \[\frac{1}{g(x)}\cdot g^{\prime}(x)=\frac{1}{y}\cdot y^{\prime}=\ln(b)\]. Now we can find that \[y^{\prime}=y\ln(b)=b^{x}\ln(b)\].

More generally, let \[y=b^{f(x)}\], \[\ln(y)=f(x)\cdot\ln(b)\]. Similar to what we've done above \[(\ln(y))^{\prime}=(f(x)\cdot\ln(b))^{\prime}\], \[\frac{1}{y}\cdot y^{\prime}=f^{\prime}(x)\cdot\ln(b)\]. Now we find that \[y^{\prime}=y\cdot f^{\prime}(x)\cdot\ln(b)=b^{f(x)}f^{\prime}(x)\ln(b)\].