Derivatives of Exponentials and Logarithms (College Board AP® Calculus AB): Revision Note

Derivative of the exponential function

How do I differentiate the exponential function?

• It can be shown that:

  • $\underset{h\to 0}{\lim }\frac{e^{x+h}-e^x}{h}=e^x$ for all real $x$

  • $\underset{x\to a}{\lim }\frac{e^x-e^a}{x-a}=e^a$ for all real $a$

• This means that $f\left(x\right)=e^x$ is an important function because its rate of change is equal to itself

  • $f^{\text{'}}\left(x\right)=e^x$

• For the function $g\left(x\right)=e^{kx}$, its rate of change is proportional to itself

  • $g^{\text{'}}\left(x\right)=k{e}^{kx}$

  • This occurs as a result of applying the chain rule

• If there is a constant multiple of the exponential, the same approach used for powers of $x$ can be applied

  • If $h\left(x\right)=a{e}^{kx}$ then $h^{\text{'}}\left(x\right)=ak{e}^{kx}$

How do I differentiate a number raised to the power of x?

• For a positive constant raised to the power of $x$,

  • If $f\left(x\right)=a^x$ where $a>0$ then $f^{\text{'}}\left(x\right)=a^x\ln a$

• This can be shown by using the identity $a^x=e^{x\ln a}$

  • $f\text{'}\left(x\right)=e^{x\ln a}·\ln a=a^x\ln a$

• If the power is a multiple of $x$,

  • $g\left(x\right)=a^{kx}$

  • $g^{\text{'}}\left(x\right)=a^{kx}k\ln a$

  • This occurs as a result of applying the chain rule

Derivative of the natural logarithmic function

How do I differentiate a natural logarithm?

• For a natural logarithm, it can be shown using inverse functions that

  • If $f\left(x\right)=\ln x$ for $x>0$ then $f^{\text{'}}\left(x\right)=\frac{1}{x}$

• This means that:

  • $\underset{h\to 0}{\lim }\frac{\ln \left(x+h\right)-\ln x}{h}=\frac{1}{x}$ for all $x>0$

  • $\underset{x\to a}{\lim }\frac{\ln x-\ln a}{x-a}=\frac{1}{a}$ for all $a>0$

• If there is a constant multiple of the logarithm, the same approach used for powers of $x$ can be applied

  • If $h\left(x\right)=a\ln x$ then $h^{\text{'}}\left(x\right)=a\left(\frac{1}{x}\right)=\frac{a}{x}$

• If there is a constant multiple of $x$ inside the logarithm,

  • $g\left(x\right)=\ln kx$

  • This can be rewritten using the laws of logarithms

    • $g\left(x\right)=\ln k+\ln x$

  • $\ln k$ is a constant, which means it has a derivative of zero

  • Therefore $g^{\text{'}}\left(x\right)=\frac{1}{x}$