Combining Exponential and Logarithmic Derivatives
Exponential and logarithmic functions often appear together in a single expression. Differentiating such an expression is usually not a matter of finding one new rule. It is a matter of identifying the outer structure of the expression, applying the product or quotient rule at that level, and then using the chain rule to differentiate each factor.
The main challenge is keeping the pieces organized. A product has two derivative contributions; a quotient has a specific numerator order; and a composite exponential or logarithm contributes an inner derivative. The examples below make each of these steps visible.
The Rules Used in This Workshop
Let \(u\) be differentiable. The chain rule formulas for exponential and logarithmic functions are:
where the second formula requires a constant base \(a>0\). For logarithms, the corresponding formulas are
where \(b>0\) and \(b\ne 1\). These conditions ensure that the real-valued functions in question are defined.
For differentiable functions \(f\) and \(g\), the product rule and quotient rule are
with the quotient rule requiring \(g(x)\ne 0\). In a mixed expression, these rules govern the outer product or quotient; the exponential and logarithmic formulas then differentiate the individual factors.
Why the Product and Quotient Rules Work
The product rule follows directly from the definition of a derivative. Suppose \(f\) and \(g\) are differentiable at \(x\). Add and subtract \(f(x+h)g(x)\) in the difference quotient:
As \(h\) approaches zero, differentiability of \(f\) implies continuity of \(f\), so \(f(x+h)\) approaches \(f(x)\). The difference quotients approach \(g'(x)\) and \(f'(x)\), respectively. Therefore, the limit is \(f(x)g'(x)+g(x)f'(x)\), which is the product rule.
For the quotient rule, assume \(g(x)\ne 0\). Near \(x\), the reciprocal \(1/g\) is defined, and the chain rule gives \((1/g)'=-g'/g^2\). Since \(f/g=f(1/g)\), applying the product rule yields
This derivation explains both the order of the two terms in the quotient-rule numerator and the square in its denominator.
Worked Example: A Natural Exponential Times a Natural Logarithm
Worked Example: A Natural Exponential Times a Natural Logarithm
Differentiate \(y=e^{x^2}\ln(2x+3)\), and state its domain.
The logarithm requires \(2x+3>0\), so \(x>-\frac{3}{2}\). The exponential is defined for every real \(x\). Set \(f(x)=e^{x^2}\) and \(g(x)=\ln(2x+3)\). The chain rule gives
Apply the product rule, retaining both contributions:
Check at \(x=0\), which lies in the domain. The factors have values \(f(0)=1\), \(f'(0)=0\), \(g(0)=\ln 3\), and \(g'(0)=2/3\). Thus the product rule gives \(y'(0)=0\cdot\ln 3+1\cdot 2/3=2/3\). Substituting \(x=0\) in the derived formula also gives \(1(0+2/3)=2/3\), as required.
Worked Example: A General-Base Exponential Times a Logarithm
Worked Example: A General-Base Exponential Times a Logarithm
Differentiate \(y=2^{3x-1}\log_5(x+2)\), and give the real domain.
The logarithm requires \(x+2>0\), so \(x>-2\). The exponential has positive constant base \(2\), so it is defined for all real \(x\). Differentiate the first factor, including the derivative of its exponent:
For the second factor, the inner function \(x+2\) has derivative \(1\), so
Now use the product rule:
As a check, evaluate both the factors and their derivatives at \(x=0\). They are \(2^{-1}=1/2\), \(3\ln(2)/2\), \(\log_5(2)\), and \(1/(2\ln(5))\), respectively. The product rule therefore gives
Substitution of \(x=0\) into the factored derivative gives the same two terms, confirming the chain factors and product-rule contributions.
Worked Example: A Logarithm Divided by a Sum of Exponentials
Worked Example: A Logarithm Divided by a Sum of Exponentials
Differentiate \(y=\dfrac{\ln(x+1)}{3^x+e^x}\), and identify its domain.
The numerator requires \(x+1>0\), or \(x>-1\). The denominator is a sum of positive terms, so it cannot be zero. Let \(f(x)=\ln(x+1)\) and \(g(x)=3^x+e^x\). Their derivatives are
Apply the quotient rule in the order \(f'g-fg'\):
Check at \(x=0\). The numerator function is \(f(0)=\ln(1)=0\), its derivative is \(f'(0)=1\), and the denominator function is \(g(0)=1+1=2\). Its derivative is \(g'(0)=\ln(3)+1\). Hence the quotient rule gives
In the displayed derivative, substituting \(x=0\) gives the same value: its numerator is \(2-0=2\), and its denominator is \(4\).
Worked Example: A Product in the Numerator and an Exponential Denominator
Worked Example: A Product in the Numerator and an Exponential Denominator
Differentiate \(y=\dfrac{e^{2x}\log_3(x+1)}{2^{x+1}}\), and state its domain.
The logarithm requires \(x+1>0\), so the domain is \(x>-1\). First differentiate the numerator as a product. Its two factors have derivatives
Thus, writing \(N(x)=e^{2x}\log_3(x+1)\) and \(D(x)=2^{x+1}\), we obtain
Using the quotient rule and then factoring out the common positive factor gives
Check at \(x=0\). The numerator is \(N(0)=1\cdot\log_3(1)=0\), its derivative is \(N'(0)=1/\ln(3)\), and \(D(0)=2\). Therefore the quotient rule gives \(y'(0)=\bigl((1/\ln(3))\cdot2-0\cdot D'(0)\bigr)/2^2=1/(2\ln(3))\). The final formula at \(x=0\) also gives \(\frac12(0+1/\ln(3))=1/(2\ln(3))\).
A Common Mistake: Losing an Inner Derivative
A frequent error is to use the correct logarithm formula but omit the derivative of its argument. For example, the incorrect calculation
does not account for the inner derivative \(\frac{d}{dx}(2x+1)=2\). The chain rule requires that factor:
For a mixed product, another common error is to differentiate only one factor. In \(f(x)g(x)\), the derivative is \(f'(x)g(x)+f(x)g'(x)\), not just \(f'(x)g(x)\). In a quotient, reversing the terms \(f'g-fg'\) also changes the sign of the result. Writing down the outer rule before simplifying helps prevent both mistakes.
A Reliable Order of Work
When an expression combines these functions, use the following sequence. It separates structural choices from the smaller chain-rule calculations.
Logarithm arguments must be positive, and a quotient denominator must be nonzero.
Decide whether the full expression is a product or a quotient, and name its factors or numerator and denominator.
For every exponential or logarithm, include the derivative of its inner function.
Combine the pieces with the product or quotient rule, then test the result at a convenient point in the domain when possible.
Why These Combinations Matter
Products and quotients of exponential and logarithmic functions occur naturally when a model combines different types of change. For example, a quantity might involve an exponential growth factor multiplied by a logarithmic response, or a logarithmic numerator scaled by an exponential denominator. Their derivatives describe how the combined expression changes, rather than how any one factor changes in isolation.
The key is to preserve the structure of the original expression long enough to apply the right rule. Simplifying too early can hide a product or quotient, while omitting a chain factor can make an otherwise correct calculation wrong. Careful notation, domain checks, and a final substitution check make these derivatives more reliable.
Check Your Understanding
For each question, identify the outer rule before carrying out the differentiation.
- Differentiate \(e^{x^2}\log_2(x+4)\), and state its real domain.
- Differentiate \(\dfrac{5^{2x}}{\ln(x+3)}\), and state the condition on \(x\).
- In differentiating \(e^{x}\ln(4x+1)\), what factor comes from the inner function of the logarithm?
- For \(f(x)/g(x)\), which order of terms appears in the numerator of the quotient rule?
- Why must the denominator of a quotient be nonzero when using the quotient rule?
