Why the Base Changes the Derivative
The functions \(2^x\), \(5^x\), and \(e^x\) all grow exponentially, but they do not grow at the same rate. Their derivatives reflect that difference: the base affects not only the value of the function but also the factor multiplying it in the derivative.
The natural exponential \(e^x\) has the special property that its derivative is itself. For a different positive base, the derivative is still proportional to the original exponential, but the proportionality factor is \(\ln a\), where \(a\) is the base. This tutorial derives that factor and uses the result in calculations.
Writing an Exponential with Base \(a\) Using \(e\)
For a positive real number \(a\), \(\ln a\) is the real number whose natural exponential is \(a\). Thus \(e^{\ln a}=a\). Using the exponent rule \((b^r)^s=b^{rs}\) for positive \(b\), we can rewrite \(a^x\):
This identity is useful because it expresses every positive-base exponential in terms of the natural exponential. Here \(a\) is fixed, so \(\ln a\) is a constant. We will use the identity to derive the derivative rather than treat \(a^x\) as a power with a fixed exponent.
Deriving the Formula
We use the definition of a derivative and the known natural-exponential limit \(\lim_{u\to 0}(e^u-1)/u=1\), which follows from the derivative of \(e^x\) at \(x=0\). First suppose \(a\ne 1\), so \(\ln a\ne 0\).
Now rewrite \(a^h=e^{h\ln a}\). Set \(u=h\ln a\). As \(h\to 0\), \(u\to 0\), and \(h=u/\ln a\). Therefore,
Substituting this limit into the difference quotient gives the formula for \(a\ne 1\). If \(a=1\), then \(1^x=1\) for every \(x\), so its derivative is \(0\); the formula also gives \(1^x\ln 1=1\cdot 0=0\). Thus the result holds for every positive base.
The proof shows where each factor comes from. The factor \(a^x\) remains when the difference quotient is factored, and the limiting change in \(a^h\) contributes \(\ln a\). In particular, the derivative is not generally just the original function.
Worked Examples
Worked Example: Differentiate \(3^x\) and Find Its Slope at \(x=2\)
Let \(f(x)=3^x\). The base \(3\) is a positive constant, so the theorem applies directly:
To find the slope at \(x=2\), substitute \(x=2\) into the derivative:
The exact slope is \(9\ln 3\). Since \(\ln 3\approx 1.098612\), the slope is approximately \(9.887508\), or \(9.888\) to three decimal places. As a check, the derivative formula at \(x=2\) is \(3^2\ln 3\), and evaluating the function first gives \(f(2)=9\), so the same slope is \(f(2)\ln 3=9\ln 3\).
Worked Example: Differentiate an Exponential with a Reciprocal Base
Let \(g(x)=(1/4)^x\). The base is positive, so apply the formula:
Because \(\ln(1/4)=-\ln 4\), this can also be written as
At \(x=1\), the function value is \(g(1)=1/4\), so
The negative sign is expected: \(0<1/4<1\), so \((1/4)^x\) decreases as \(x\) increases. Checking by substituting \(x=1\) into the derivative expression gives \((1/4)^1\ln(1/4)=-\ln(4)/4\), matching the result.
Worked Example: Differentiate a Constant Multiple of an Exponential
Let \(h(x)=6\cdot 5^x\). The constant-multiple rule says that the derivative of a constant times a differentiable function is that constant times the function's derivative. First apply the exponential formula, then keep the factor \(6\):
At \(x=0\), \(5^0=1\), so
The check is direct: substituting \(x=0\) into the derivative found above gives \(6\cdot 5^0\ln 5=6\ln 5\), as required. The outside constant \(6\) remains in the derivative; it does not become part of the exponent.
What the Sign and Size of \(\ln a\) Tell You
The derivative formula also helps interpret how an exponential changes. The function \(a^x\) is positive for every real \(x\) when \(a>0\), so the sign of its derivative is the sign of \(\ln a\).
| Base | Sign of \(\ln a\) | Behavior of \(a^x\) |
|---|---|---|
| \(a>1\) | Positive | The derivative is positive, so the function increases. |
| \(a=1\) | Zero | The function is constant, and its derivative is zero. |
| \(0<a<1\) | Negative | The derivative is negative, so the function decreases. |
For example, \(5^x\) has positive derivative because \(\ln 5>0\), whereas \((1/4)^x\) has negative derivative because \(\ln(1/4)<0\). This conclusion follows from the formula and the fact that \(a^x>0\); it does not require evaluating the function at many inputs.
Why the Positive-Base Condition Matters
The condition \(a>0\) ensures that \(a^x\) is a real-valued function for every real \(x\), and that \(\ln a\) is defined as a real number. A negative base does not give a real value for every real exponent. For instance, \((-2)^{1/2}\) is not real. Therefore the formula is not a derivative rule for negative bases as functions defined on all real inputs.
The base \(a=1\) is allowed, but it is a special constant-function case. A base of zero is not included either: \(0^x\) is not defined for every real \(x\), and the expression \(\ln 0\) is not defined.
Common Mistake: Using the Power Rule
A frequent error is to treat \(a^x\) as if it were a power \(x^n\) with a fixed exponent. For example, the incorrect calculation
uses the power rule in the wrong situation. The power rule differentiates a variable base raised to a constant exponent, such as \(x^3\). In \(2^x\), the base is constant and the exponent varies. The correct calculation is
Another version of the error is to write \((2^x)'=2^x\), forgetting \(\ln 2\). The natural exponential is the special case \(a=e\), for which \(\ln e=1\). For \(a=2\), the factor is \(\ln 2\), not \(1\). At \(x=0\), the mistaken formula predicts slope \(1\), while the correct formula gives \(\ln 2\).
Check Your Understanding
Use the derivative formula for a positive constant base. Keep answers exact unless a question asks for an approximation.
- Find the derivative of \(7^x\).
- Find the slope of \(4^x\) at \(x=1\).
- Differentiate \((1/3)^x\), and state whether its derivative is positive or negative.
- Find the derivative of \(9\cdot 2^x\).
- Explain why the formula \(\frac{d}{dx}a^x=a^x\ln a\) does not apply to a negative base as a real-valued function on all real inputs.
