
Exponential and logarithmic functions play a crucial role in calculus because they describe growth, decay, and scaling processes found in science, economics, and engineering.
In this lecture, you’ll learn how to find derivatives of both exponential functions and logarithmic functions, and how they relate to the natural base eee.
The most important exponential function is:
$$f(x)=e^x$$
Its derivative is special because it’s the same as the original function:
$$f^\prime(x)=\frac{d}{dx}[e^x]=e^x$$
👉 The rate of change of $$e^x$$ equals its value — this is why it’s so useful in modeling natural growth.
For a general exponential function:
$$f(x)=a^x (a>0 , a\neq0) =e^{ln{a^{x}}}=e^{x \cdot ln(a)}$$
the derivative is:
$$f^\prime(x)=e^{x \cdot ln(a)} \cdot ln(a) =e^{ln(a)^x} \cdot ln(a) =a^{x} \cdot ln(a)$$
$$f^\prime(x)=\frac{d}{dx}[a^x]=a^{x} \cdot ln(a)$$
✅ Example:
$$\frac{d}{dx}[2^x]=2^xln(2)$$
When the exponent is more than just xxx, we use the Chain Rule.
$$f(x)=e^{u(x)}$$
then
$$f^\prime(x)=\frac{d}{dx}[e^{u(x)}]=e^{u(x)}\cdot u^\prime(x)$$
and
$$f(x)=a^{u(x)}$$
then
$$f^\prime(x)=\frac{d}{dx}[a^{u(x)}]=ln(a)a^{u(x)}\cdot u^\prime(x)$$
✅ Example:
$$\frac{d}{dx}[e^{3x}]=e^{3x}\cdot(3x)^\prime=3e^{3x}$$
$$\frac{d}{dx}[ln(x)]=\frac{1}{x}$$
For $$f(x)=log_{a}(x)$$
$$f^\prime(x)=\frac{d}{dx}=\frac{1}{xln(a)}$$
✅ Example:
$$\frac{d}{dx}[log_{2}(x)]=\frac{1}{xln(2)}$$
$$f(x)=ln(u(x))$$
then by the Chain Rule:
$$f^\prime(x)=\frac{d}{dx}[ln(u(x))]=\frac{1}{u(x)}\frac{du}{dx}=\frac{1}{u(x)}\cdot u^\prime(x)$$
and
$$f(x)=log_{a}(u(x))$$
then by the Chain Rule:
$$f^\prime(x)=\frac{d}{dx}[log_{a}(u(x))]=\frac{1}{u(x)ln(a)}\frac{du}{dx}$$
✅ Example:
$$\frac{d}{dx}[log_{3}(2x^2+5x+1)]=\frac{1}{(2x^2+5x+1)ln(3)}\cdot(2x^2+5x+1)^\prime=\frac{4x+5}{(2x^2+5x+1)ln(3)}$$
Forgetting to multiply by the derivative of the inner function u′(x)u'(x)u′(x).
Using log(x) instead of ln(x) without considering the base.
Dropping ln(a) when differentiating $$a^{x}$$.
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