Linear Exponents and Instantaneous Rates
A natural exponential can change rapidly even when its exponent is a simple linear expression. For example, in \(A e^{kx+b}\), the constants \(A\), \(k\), and \(b\) affect the function in different ways. When we differentiate with respect to \(x\), the coefficient \(k\) in the exponent becomes an essential factor in the instantaneous rate.
This matters whenever a model describes a quantity that changes exponentially with time or another measured input. The derivative gives the function’s instantaneous rate of change at a particular input. Its sign tells us whether the function is increasing or decreasing there, and its magnitude gives the steepness of its graph.
Why the Coefficient \(k\) Appears
The chain rule gives a quick justification: the derivative of \(e^{u}\) with respect to \(u\) is \(e^{u}\), and the derivative of \(kx+b\) with respect to \(x\) is \(k\). Multiplying by the constant \(A\), the chain rule gives \(Ak e^{kx+b}\).
The same result follows directly from the definition of the derivative. This derivation shows explicitly how a change in the exponent contributes the factor \(k\). We use the standard limit \(\lim_{z\to 0}(e^z-1)/z=1\).
If \(k\ne 0\), set \(z=kh\). As \(h\to 0\), \(z\to 0\), and \(h=z/k\). Therefore,
Thus \(f'(x)=Ak e^{kx+b}\) when \(k\ne 0\). If \(k=0\), then \(f(x)=A e^b\) is constant, so \(f'(x)=0\), which agrees with the same formula. This establishes the derivative rule for all real values of \(A\), \(k\), and \(b\).
Applying the Rule at a Particular Input
To find an instantaneous rate at a specified input, first differentiate the whole function, then substitute the input into the derivative. Keeping those steps separate makes it easier to include the chain factor and evaluate the exponent correctly.
Worked Example: Finding a Slope at a Given Input
Let \(f(x)=3e^{2x-1}\). Find the instantaneous rate of change at \(x=1\).
Here \(A=3\), \(k=2\), and \(b=-1\). Applying the rule gives
Substitute \(x=1\) into the derivative:
For a check, the function value at that input is \(f(1)=3e^{2-1}=3e\), and the derivative formula evaluated there is \(6e^{2-1}=6e\). Thus the requested instantaneous rate is \(6e\), approximately \(16.31\) if rounded to two decimal places.
How the Sign of the Function Affects the Rate
The exponential factor is positive, but \(A\) need not be positive. Therefore, it is not enough to look at \(k\) alone to decide whether the function is increasing or decreasing. The sign of \(Ak\) determines the sign of \(f'(x)\) at every input.
- If \(Ak>0\), then \(f'(x)>0\), so the function is increasing.
- If \(Ak<0\), then \(f'(x)<0\), so the function is decreasing.
- If \(Ak=0\), then \(f'(x)=0\) for every \(x\), and the function is constant.
Worked Example: A Negative Function with a Positive Rate
Let \(g(t)=-4e^{-3t+2}\). Find \(g'(0)\), and determine whether the function is increasing or decreasing.
The constants are \(A=-4\) and \(k=-3\). Differentiate:
At \(t=0\), the exponent is \(-3(0)+2=2\), so
Because \(12e^{-3t+2}>0\) for every \(t\), the function is increasing. In particular, \(g(0)=-4e^2\) is negative while \(g'(0)=12e^2\) is positive: the function’s value is negative, but it is increasing at that input. Substituting \(t=0\) into the derivative expression confirms the rate \(12e^2\), approximately \(88.67\) to two decimal places.
What the Parameters Change
The formula separates the roles of the constants. The factor \(A\) scales the function and its derivative, including reversing their signs if \(A\) is negative. The coefficient \(k\) affects the derivative through its sign and size. The constant \(b\) occurs in the exponential factor and changes the function’s value and slope at a given input.
These effects should be interpreted at a specified input when comparing actual rates. Even if two models have different values of \(k\), their slopes also depend on their other constants and on the input being considered. The rule supplies the exact derivative for each model; it does not say that \(k\) alone determines every numerical slope.
Worked Example: Interpreting the Rate and Its Units
Suppose a quantity measured in units \(Q\) is modeled, with time \(t\) in hours, by \(P(t)=120e^{0.2t-1}\). Find the instantaneous rate at \(t=5\), and state the units of the rate.
Differentiate with respect to \(t\), including the factor \(0.2\):
At \(t=5\), the exponent is \(0.2(5)-1=1-1=0\). Therefore,
The model value at that time is \(P(5)=120e^0=120\) units \(Q\). The rate is \(24\) units \(Q\) per hour, because the derivative measures change in \(P\) per unit change in time. Checking the derivative expression at \(t=5\) again gives \(24e^0=24\), as required.
Units in the Exponent and the Derivative
An exponential’s exponent must be dimensionless. If \(t\) is measured in hours, then \(kt\) can be dimensionless only when \(k\) has units of inverse hours. The constant \(b\) is dimensionless. If \(A\) has units of \(Q\), then \(A e^{kt+b}\) has units of \(Q\), while \(Ak e^{kt+b}\) has units of \(Q\) per hour.
This unit check is useful for detecting errors in applied models. A derivative with respect to time should have units of quantity per time. If the factor \(k\) is missing, the units will generally expose the problem: the result would have the units of the quantity, not the units of a rate.
Special Cases and a Common Pitfall
If \(k=0\), the exponent does not vary with \(x\), and \(A e^{kx+b}=A e^b\) is constant. Its derivative is zero. If \(A=0\), the function is also constant at zero, regardless of \(k\). Both cases are included in the formula \(Ak e^{kx+b}\).
A frequent mistake is to differentiate the exponential factor but omit the derivative of its exponent. The chain rule requires that inner derivative as a multiplier.
Worked Example: Correcting a Missing Chain Factor
For \(q(t)=7e^{4t-2}\), an incorrect calculation might be
This omits the derivative of \(4t-2\), which is \(4\). Applying the chain rule correctly gives
At \(t=0\), the incorrect result would give \(7e^{-2}\), whereas the correct rate is
For a check, differentiating the original expression again by the rule gives \(7\cdot 4e^{4t-2}=28e^{4t-2}\). The missing factor changes the rate by a factor of four, so it cannot be dropped.
A Reliable Calculation Process
For a function with a constant multiple and a linear exponent, organize the work as follows:
Write down the coefficient \(A\) and the exponent \(kx+b\), so the inner coefficient \(k\) is visible.
Use \(f'(x)=Ak e^{kx+b}\), retaining the same exponent and including \(k\).
Substitute the requested input into the derivative, simplify the exponent, and then evaluate the exponential.
Use the sign of \(Ak\) to determine the direction of change, and confirm the units are output units per input unit when the variables have units.
The main result is compact, but its interpretation depends on keeping the full expression in view. The coefficient \(k\) enters because the exponent changes at rate \(k\); the other factors and the input determine the remaining value of the slope.
Check Your Understanding
For each question, show enough work to make the chain factor and the evaluation clear.
- Differentiate \(r(x)=5e^{-2x+3}\).
- Find the instantaneous rate of change of \(s(x)=-2e^{3x-1}\) at \(x=0\). Is the function increasing or decreasing?
- A model is \(M(t)=50e^{0.1t+2}\), where \(t\) is measured in days and \(M\) in grams. Find the rate at \(t=0\), including its units.
- What is the derivative of \(u(x)=9e^4\), and why does the general formula give the same result?
- For \(v(x)=6e^{kx+b}\), which constants determine the sign of \(v'(x)\), and why can the exponential factor be excluded from that sign check?
