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.