Why the Base Changes the Derivative
The natural exponential \(e^x\) has the special property that its derivative is itself. For another positive base, the derivative is still proportional to the original exponential, but the proportionality factor depends on the base.
That factor is \(\ln a\). To see why, we express \(a^x\) using the natural exponential and then apply the chain rule. This representation also makes clear why the derivative can be positive, negative, or zero, depending on the base.
Writing \(a^x\) as a Natural Exponential
Let \(a>0\) be a fixed real number. The natural logarithm \(\ln a\) is defined for every positive \(a\), and \(e^{\ln a}=a\). Consequently, for any real \(x\),
Here \(a\) is constant, so \(\ln a\) is also constant. The exponent \(x\ln a\) is a constant multiple of \(x\). Differentiating this expression will therefore use the derivative of the natural exponential and the chain rule.
Deriving the Derivative Formula
The derivative of \(e^u\) with respect to \(u\) is \(e^u\). By the chain rule, if \(u=u(x)\), then the derivative of \(e^{u(x)}\) is \(e^{u(x)}u'(x)\). In the expression \(a^x=e^{x\ln a}\), the inner function is \(u(x)=x\ln a\). Since \(\ln a\) is constant, \(u'(x)=\ln a\).
The first equality uses the representation of \(a^x\) as a natural exponential. The second applies the chain rule, and the third differentiates a constant multiple of \(x\). The final equality returns to the original base.
The condition \(a>0\) ensures that \(\ln a\) exists as a real number and that \(a^x\) is defined as a real-valued exponential function for every real \(x\). The base is held fixed while \(x\) varies. This result concerns a constant base and the exponent \(x\); it does not by itself differentiate expressions with a variable or composite exponent.
What the Formula Says About Different Bases
The sign of \(\ln a\) determines the sign of the derivative because \(a^x>0\) for every real \(x\). If \(a>1\), then \(\ln a>0\), so \(a^x\) has positive derivative. If \(0<a<1\), then \(\ln a<0\), so its derivative is negative. For \(a=1\), the derivative is zero.
| Base | Sign of \(\ln a\) | Sign of \(\frac{d}{dx}a^x\) |
|---|---|---|
| \(a>1\) | Positive | Positive |
| \(0<a<1\) | Negative | Negative |
| \(a=1\) | Zero | Zero |
The factor \(\ln a\) is the rate multiplier that distinguishes one base from another. For \(a=e\), the multiplier is \(\ln e=1\), recovering the familiar derivative of \(e^x\). For other bases it is generally not one.
Worked Examples
Worked Example: Differentiate \(2^x\) and find its slope at \(x=3\)
Here the base is \(a=2\), which is positive. Apply the formula:
At \(x=3\), first evaluate the exponential: \(2^3=8\). Thus the slope is
Check by differentiating the rewritten form. Since \(2^x=e^{x\ln 2}\),
At \(x=3\), this is \(8\ln 2\), agreeing with the result.
Worked Example: Differentiate \(\left(\frac14\right)^x\) and evaluate at \(x=-1\)
The base \(a=\frac14\) is positive and less than one. The derivative function is
The second line uses \(\ln(1/4)=-\ln 4\). At \(x=-1\), the exponential value is
Therefore the slope is
Check using \(\left(\frac14\right)^x=4^{-x}=e^{-x\ln 4}\). Differentiation gives
At \(x=-1\), \(4^{-(-1)}=4\), so the derivative is \(-4\ln 4\), as before. Its negative sign is consistent with a base between zero and one.
Worked Example: Differentiate \(1^x\)
The general formula applies because \(1>0\). Substituting \(a=1\) gives
The same conclusion follows directly from \(1^x=1\): the function is constant, and its derivative is zero. Check the formula at, for example, \(x=5\). Since \(1^5=1\), the formula gives \((\ln 1)1^5=0\cdot 1=0\), matching the derivative of the constant function.
Worked Example: Differentiate \(e^x\) using the general-base formula
Take \(a=e\). Since \(\ln e=1\), the formula yields
For a check at \(x=0\), the formula gives \(e^0=1\). Differentiating the function \(e^x\) by the natural-exponential rule also gives \(e^x\), which equals \(1\) at \(x=0\).
Why the Derivation Works
The essential point is that the base \(a\) can be encoded as a natural exponential: \(a=e^{\ln a}\). Raising that equality to the power \(x\) makes the exponent \(x\ln a\). The dependence on \(x\) is now visible as a simple inner function, and the chain rule accounts for its derivative.
Use \(a=e^{\ln a}\), which is valid for every \(a>0\).
Obtain \(a^x=e^{x\ln a}\), with inner function \(x\ln a\).
Differentiate the outer exponential and multiply by the inner derivative \(\ln a\).
Replace \(e^{x\ln a}\) with \(a^x\) to obtain \((\ln a)a^x\).
This also explains the special case \(a=1\) without needing a separate exception to the theorem. The rewrite gives \(1^x=e^{x\ln 1}=e^0=1\), and the inner derivative is \(\ln 1=0\). Hence the chain rule gives zero, exactly as differentiation of a constant requires.
Common Mistake: Omitting the Factor \(\ln a\)
A frequent error is to treat every exponential as though its derivative were itself. That property holds for \(e^x\), not for every base. For example, the incorrect calculation
omits the change in the exponent when \(5^x\) is rewritten as \(e^{x\ln 5}\). The corrected calculation is
The omitted factor is not generally equal to one: \(\ln 5\ne 1\). A quick check is to set the base to \(1\). The incorrect rule would give the derivative of \(1^x\) as \(1\), even though \(1^x\) is the constant function \(1\) and its derivative is \(0\). The correct formula gives \((\ln 1)1^x=0\).
Another condition to keep in view is that the base must be positive for this real-variable formula. If \(a\leq 0\), then \(\ln a\) is not defined as a real logarithm, and \(a^x\) is not a real-valued function for every real \(x\). The theorem therefore makes no general claim for such bases.
Why This Result Matters
The derivative formula provides a direct way to measure how a positive-base exponential changes at any input. It also shows that exponential functions retain their original form under differentiation: only the constant multiplier \(\ln a\) is added. That multiplier explains the direction of change and distinguishes growth for bases above one from decay for bases between zero and one.
The proof is useful beyond this one formula because it illustrates a general strategy: rewrite an expression using a function whose derivative is already known, then use the chain rule to account for the inner expression. Here, the natural exponential is the convenient outer function, while \(x\ln a\) supplies the required factor.
Check Your Understanding
Use the conditions and derivative formula above to answer the following questions.
- For a fixed positive base \(a\), write \(a^x\) as a natural exponential and identify the derivative of its exponent.
- Find the derivative of \(7^x\). Explain why the answer is not simply \(7^x\).
- Find the derivative of \(\left(\frac13\right)^x\) and state whether it is positive or negative for every real \(x\).
- Use the general formula to differentiate \(1^x\), then explain why this agrees with the fact that \(1^x\) is constant.
- State the condition on \(a\) required for the theorem and explain why the formula is not stated for every real base.
- For what positive base does the formula reduce to \(\frac{d}{dx}a^x=a^x\), and why?
