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3.9 Derivatives of Exponential and Logarithmic Functions

3.9 Derivatives of Exponential and Logarithmic Functions

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
Calculus I
Unit & Topic Study Guides

Exponential and logarithmic functions show up constantly in calculus because they model growth, decay, and many natural processes. Their derivatives follow clean, predictable rules, and once you know them, a huge range of problems becomes manageable.

Exponential and Logarithmic Differentiation

Differentiation of exponential functions

The core rule: the derivative of axa^x is axln(a)a^x \ln(a), where aa is a positive constant not equal to 1. You're multiplying the original function by the natural log of its base.

  • ddxax=axln(a)\frac{d}{dx}a^x = a^x \ln(a)
  • Example: ddx2x=2xln(2)\frac{d}{dx}2^x = 2^x \ln(2)

The natural exponential function exe^x is a special case. Since ln(e)=1\ln(e) = 1, the derivative of exe^x is just exe^x. This is the only function that is its own derivative, which is a big part of why ee is so important in calculus.

  • ddxex=ex\frac{d}{dx}e^x = e^x

Constant multiples work the way you'd expect:

  • ddx(kax)=kaxln(a)\frac{d}{dx}(k \cdot a^x) = k \cdot a^x \ln(a)
  • ddx(5ex)=5ex\frac{d}{dx}(5e^x) = 5e^x

When the exponent is a function u(x)u(x) rather than just xx, apply the chain rule by multiplying by u(x)u'(x):

  • ddxeu(x)=eu(x)u(x)\frac{d}{dx}e^{u(x)} = e^{u(x)} \cdot u'(x)
  • ddxau(x)=au(x)ln(a)u(x)\frac{d}{dx}a^{u(x)} = a^{u(x)} \cdot \ln(a) \cdot u'(x)
  • Example: ddxesinx=esinxcosx\frac{d}{dx}e^{\sin x} = e^{\sin x} \cdot \cos x
Differentiation of exponential functions, Derivatives and the Shape of a Graph · Calculus

Derivatives of logarithmic functions

The derivative of loga(x)\log_a(x) is 1xln(a)\frac{1}{x \ln(a)}, where aa is a positive constant not equal to 1 and x>0x > 0.

  • ddxloga(x)=1xln(a)\frac{d}{dx}\log_a(x) = \frac{1}{x \ln(a)}
  • Example: ddxlog2(x)=1xln(2)\frac{d}{dx}\log_2(x) = \frac{1}{x \ln(2)}

For the natural logarithm, ln(a)=ln(e)=1\ln(a) = \ln(e) = 1, so the formula simplifies:

  • ddxln(x)=1x\frac{d}{dx}\ln(x) = \frac{1}{x}, where x>0x > 0

Constant multiples carry through as usual:

  • ddx(kln(x))=kx\frac{d}{dx}(k \cdot \ln(x)) = \frac{k}{x}
  • Example: ddx(3ln(x))=3x\frac{d}{dx}(3\ln(x)) = \frac{3}{x}

When the argument is a function u(x)u(x), apply the chain rule. This gives you u(x)u'(x) in the numerator:

  • ddxln(u(x))=u(x)u(x)\frac{d}{dx}\ln(u(x)) = \frac{u'(x)}{u(x)}
  • ddxloga(u(x))=u(x)u(x)ln(a)\frac{d}{dx}\log_a(u(x)) = \frac{u'(x)}{u(x) \ln(a)}
  • Example: ddxln(cosx)=sinxcosx=tanx\frac{d}{dx}\ln(\cos x) = \frac{-\sin x}{\cos x} = -\tan x

A common mistake: forgetting the chain rule when the argument isn't just xx. If you see ln(something)\ln(\text{something}), always ask yourself whether that "something" needs its own derivative.

Differentiation of exponential functions, derivatives - What is the difference between exponential symbol $a^x$ and $e^x$ in mathematics ...

Logarithmic differentiation techniques

Some functions are difficult or impossible to differentiate with standard rules alone. Logarithmic differentiation handles these by taking ln\ln of both sides and then using implicit differentiation.

This technique is especially useful for:

  • Functions of the form [u(x)]v(x)[u(x)]^{v(x)}, where both the base and exponent depend on xx
  • Products of many factors, like f(x)=u1(x)u2(x)u3(x)f(x) = u_1(x) \cdot u_2(x) \cdot u_3(x)

The key idea is that logarithm properties convert products into sums and exponents into coefficients, making differentiation much simpler.

Steps for logarithmic differentiation:

  1. Set y=f(x)y = f(x) and take the natural log of both sides: ln(y)=ln(f(x))\ln(y) = \ln(f(x))
  2. Use log properties to simplify: ln(ab)=ln(a)+ln(b)\ln(ab) = \ln(a) + \ln(b) and ln(ab)=bln(a)\ln(a^b) = b\ln(a)
  3. Differentiate both sides with respect to xx. The left side becomes 1ydydx\frac{1}{y} \cdot \frac{dy}{dx} by implicit differentiation
  4. Solve for dydx\frac{dy}{dx} by multiplying both sides by yy, then substitute the original expression back in for yy

Example: Find ddx(xx)\frac{d}{dx}(x^x).

  1. Let y=xxy = x^x, so ln(y)=ln(xx)\ln(y) = \ln(x^x)
  2. Simplify: ln(y)=xln(x)\ln(y) = x \ln(x)
  3. Differentiate both sides: 1ydydx=ln(x)+x1x=ln(x)+1\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1
  4. Solve: dydx=y(ln(x)+1)=xx(ln(x)+1)\frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1)

Notice that in step 3, the right side uses the product rule on xln(x)x \ln(x). That's easy to miss if you rush through it.

Advanced Techniques and Applications

Two techniques from later in the course connect directly to exponential and logarithmic functions:

  • The inverse function rule provides a way to derive the logarithmic derivative from the exponential derivative (and vice versa), since ln(x)\ln(x) and exe^x are inverses of each other.
  • L'Hôpital's rule is used to evaluate limits that produce indeterminate forms like 00\frac{0}{0} or \frac{\infty}{\infty}. Many of these limits involve exponential or logarithmic expressions, so being comfortable with their derivatives is essential when you reach that topic.
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