From the Basic Exponential to a Composite Exponential
The derivative of \(e^x\) is \(e^x\). This special property makes natural exponentials convenient to differentiate, but it does not mean that every expression involving \(e\) has derivative equal to itself. If the exponent is a function of \(x\), its rate of change must also be included.
For example, in \(e^{x^2}\), the exponent changes at a rate of \(2x\). The derivative therefore involves both the original exponential and the derivative of its exponent. This is an application of the chain rule: differentiate the outer function, then multiply by the derivative of the inner function.
The Rule and Its Conditions
Let \(I\) be an open interval, let \(u:I\to\mathbb{R}\) be differentiable, and define \(f:I\to\mathbb{R}\) by \(f(x)=e^{u(x)}\). Since the exponential function is differentiable at every real number, the chain rule applies at every \(x\in I\).
The result is local as well: it is enough for \(u\) to be differentiable at a particular point \(a\) to conclude that the composite is differentiable there and that \(f'(a)=e^{u(a)}u'(a)\). The exponent must have a real value in a neighborhood of the point under consideration, so the real-valued natural exponential is defined. In the usual real-variable setting, a differentiable real-valued \(u\) on an open interval meets this condition.
To use the formula, identify \(u(x)\), keep the entire exponent in the exponential factor, and multiply by \(u'(x)\). The derivative of \(u\) is not a replacement for the exponential factor; both factors are needed.
Why the Exponent’s Derivative Appears
The chain rule explains the multiplication in the formula. The outer function is \(g(y)=e^y\), and the inner function is \(u(x)\). The derivative of the outer function at \(y\) is \(g'(y)=e^y\). Substituting \(y=u(x)\) gives \(g'(u(x))=e^{u(x)}\), and the chain rule multiplies this by \(u'(x)\).
Here is the same reasoning from the definition of derivative. Fix a point \(a\) where \(u\) is differentiable, and write \(\Delta u=u(a+h)-u(a)\). Differentiability of \(u\) gives \(\Delta u=u'(a)h+o(h)\) as \(h\to0\). Differentiability of the exponential at \(u(a)\), together with \((e^y)'=e^y\), gives
Because \(\Delta u=u'(a)h+o(h)\), we have \(\Delta u=O(h)\). Thus the remainder \(o(\Delta u)\) is also \(o(h)\). Substituting the expression for \(\Delta u\) yields
Divide by \(h\ne0\) and take the limit as \(h\to0\). The term \(o(h)/h\) tends to zero, so
This proves the stated derivative formula at \(a\). It also shows why the exponent’s rate of change must appear: a small change in \(x\) first changes \(u(x)\), and the exponential responds to that change.
Worked Examples
Worked Example: A Polynomial Exponent
Differentiate \(f(x)=e^{x^2+1}\), then find \(f'(1)\).
Set \(u(x)=x^2+1\). Differentiate the exponent: \(u'(x)=2x\). Apply the rule to the whole exponent:
At \(x=1\), the exponent is \(1^2+1=2\), and the exponent’s derivative is \(2(1)=2\). Therefore
Check: the outer derivative contributes \(e^{x^2+1}\), and the inner derivative contributes \(2x\). Evaluating those two factors at \(1\) gives \(e^2\) and \(2\), whose product is \(2e^2\), as found.
Worked Example: A Trigonometric Exponent
Differentiate \(g(x)=e^{\sin x}\), then find \(g'\left(\frac{\pi}{6}\right)\). The sine derivative formula is applied with angles measured in radians.
Here \(u(x)=\sin x\), so \(u'(x)=\cos x\). The chain rule gives
At \(x=\frac{\pi}{6}\), \(\sin\left(\frac{\pi}{6}\right)=\frac12\) and \(\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\). Substituting both values gives
Check: the exponential factor at this input is \(e^{1/2}\), and the rate of change of the exponent is \(\sqrt{3}/2\). Their product is the derivative value above.
Worked Example: A Cubic Polynomial Exponent
Differentiate \(h(x)=e^{x^3-2x}\), then find \(h'(2)\).
Let \(u(x)=x^3-2x\). Differentiating term by term gives \(u'(x)=3x^2-2\). Therefore
At \(x=2\), evaluate the exponent and its derivative separately:
Hence
Check: substituting \(x=2\) into the derivative formula gives \(e^{2^3-2(2)}(3(2^2)-2)=e^4(10)=10e^4\), agreeing with the separate evaluation.
A Reliable Procedure
For a derivative of the form \(e^{u(x)}\), use the following steps. The important point is to treat the exponent as a complete inner function rather than as a decorative part of the notation.
Write \(u(x)\) for the complete exponent, including all its terms and operations.
Find \(u'(x)\) using the appropriate differentiation rules.
Write \(e^{u(x)}\) unchanged, then multiply it by \(u'(x)\).
If a derivative value is required, substitute the input into both factors and simplify their product.
Common Mistake: Omitting the Chain Factor
A frequent error is to use the fact that the derivative of \(e^x\) is \(e^x\), but overlook that the exponent is not simply \(x\). For \(q(x)=e^{3x^2}\), the incorrect calculation is
This calculation differentiates the outer exponential but does not account for the changing exponent. The inner function is \(u(x)=3x^2\), and its derivative is \(u'(x)=6x\). The corrected calculation is
For a concrete check at \(x=1\), the incorrect expression gives \(e^3\), whereas the rule gives \(6e^3\). The factor \(6\) is required because the exponent’s rate of change at \(1\) is \(6\). Omitting it gives the wrong rate.
A related notation error is to write \(e^{u'(x)}\). That changes the exponent rather than multiplying by its derivative. The correct structure is \(e^{u(x)}u'(x)\): the original exponent stays inside \(e^{(\cdot)}\), and its derivative is a separate factor.
When the Rule Is Useful
Natural exponentials of functions occur whenever a quantity depends exponentially on a changing input. The chain rule lets us compute the instantaneous rate of change without treating the exponent as constant. Its two factors have distinct roles: \(e^{u(x)}\) records the current exponential value, while \(u'(x)\) records how rapidly the exponent changes.
This distinction is also useful for checking an answer. If \(u'(x)=0\) at a point, the composite derivative there must be zero, because the exponent has zero instantaneous rate of change. If \(u'(x)\) is negative, then \(e^{u(x)}\) remains positive, so the derivative is negative. These conclusions follow directly from the product \(e^{u(x)}u'(x)\).
Before accepting a result, check three things: the exponent in the exponential is unchanged, the derivative of the entire exponent is present, and that derivative is multiplied rather than placed in the exponent. Also verify that the exponent is differentiable at the point where the formula is being used.
Check Your Understanding
For each question, identify the exponent and its derivative before simplifying.
- Differentiate \(f(x)=e^{x^4}\).
- Differentiate \(g(x)=e^{\cos x}\).
- For \(p(x)=e^{2x^3-x}\), find \(p'(1)\).
- If \(r(x)=e^{u(x)}\), with \(u\) differentiable and \(u'(a)=0\), what is \(r'(a)\)? Explain using the rule.
- A student claims that the derivative of \(e^{5x^2+1}\) is \(e^{5x^2+1}\). Identify the missing factor and give the corrected derivative.
