# Derivatives of logarithmic functions

# 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.