# Derivatives of logarithmic functions

• Recall that if $$y=f^{-1}(x)$$ is an inverse function, then we have $$y'=\frac{1}{f'(f^{-1}(x))}$$.22
• Let us now consider the case $$f(x)=e^x$$.
• That is, we want to compute the derivative of $$f^{-1}(x)=\ln(x)$$.
• Substituting to the formula, we get $(\ln(x))'=\frac{1}{f'(\ln(x))}=\frac{1}{e^{\ln x}}=\frac{1}{x}.$
• Exercises. 3.6: 4, 6, 8, 18, 22, 28, 30
• Consider now the function $$f(x)=\ln|x|$$.
• Its domain is $$(-\infty,0)\cup(0,\infty)$$.
• If $$x>0$$, then we have $(\ln|x|)'=(\ln(x))'=\frac{1}{x}.$
• If $$x<0$$, then we have $(\ln|x|)'=(\ln(-x))'=\frac{1}{-x}(-1)=\frac{1}{x}.$
• We've gotten that $$(\ln|x|)'=\frac{1}{x}$$ on the entire domain $$(-\infty,0)\cup(0,\infty)$$. This will be important for integration.
• Exercise. 3.6: 16.

# Logarithmic differentiation

• When having to derivate products, quotients or exponents, it might be useful to take natural logarithms of the equation $$y=f(x)$$ and use implicit differentiation.
• For example, let's prove the power rule $$(x^n)'=nx^{n-1}$$ for an arbitrary real number $$n$$.
• Note that you can use the limit definition in case $$n$$ is rational, but not in case $$n$$ is irrational.
• Taking natural logarithms in the equation $y=x^n,$ we get $\ln(y)=n\ln(x).$
• Differentiating both sides with respect to $$x$$, we get $y^{-1}y'=nx^{-1}.$
• Solving for $$y'$$, we get $y'=nyx^{-1}=nx^nx^{-1}=nx^{n-1}$ as we wanted.
• Exercises. 3.6: 40, 44, 46.

# The number $$e$$ as a limit.

• Recall that we have defined the number $$e$$ as the real number $$b$$ such that the tangent line to the graph $$y=b^x$$ at $$x=0$$ has slope 1.
• Now we'll be able to express $$e$$ as a limit.
• Letting $$f(x)=\ln(x)$$, we get $f'(x=1)=\frac{1}{1}=1.$
• On the other hand, the limit definition says $f'(x=1)=\lim_{h\to0}\frac{\ln(1+h)-\ln(1)}{h}=\lim_{h\to0}(h^{-1}\ln(1+h))=\lim_{h\to0}\ln((1+h)^{1/h}).$
• That is, we have $\lim_{x\to0}\ln((1+x)^{1/x})=1.$
• This in turn gives $\lim_{x\to0}(1+x)^{1/x}=\lim_{x\to0}e^{\ln((1+x)^{1/x})}=e^{\lim_{x\to0}\ln((1+x)^{1/x})}=e^1=e.$
• That is, we have the limit expression $\lim_{x\to0}(1+x)^{1/x}=e.$
• Note that this expression approaches $$e$$ rather slowly, as you can see in Figure 4 in Section 3.6 of the textbook. For example, to get precision up to 4 decimals, you almost need $$x<0.0001$$.
• Exercise. 3.6: 56.