From a Small Input Change to an Estimated Output
A function describes how an output depends on an input. If the input changes, the output will usually change too. When the input change is small, a derivative gives a convenient way to estimate the corresponding output change without evaluating the function at the new input exactly.
The key is to keep two quantities separate. The estimated output change tells us how much the output is predicted to change. The estimated new output adds that change to the output at the original input. Forgetting this second step is a common source of incorrect answers.
Input Increments and Differentials
Suppose the input starts at \(x\) and changes to \(x+dx\). The signed input increment is \(dx\): it is positive if the input increases and negative if the input decreases. The actual output change is
The differential \(dy\) is defined using the derivative at the starting input:
Thus, to estimate the finite change, calculate the derivative at the original input, multiply by the signed input increment, and interpret the result as an estimated output change. The new input \(x+dx\) is used when describing the change, but the slope in the differential is evaluated at \(x\).
The estimate is local: it uses the derivative at the starting point to model the function nearby. It is appropriate when \(f\) is differentiable at that point, the new input is in the function’s domain, and the increment is small enough for a local linear model to be useful. The formula does not claim that \(dy\) is exactly equal to \(\Delta y\).
Why the Differential Gives a Useful Estimate
The derivative \(f'(x)\) is the slope of the tangent line to the graph of \(f\) at the point \((x,f(x))\). A horizontal movement of \(dx\) along that tangent line produces a vertical movement of slope times horizontal change, namely \(f'(x)\,dx\). The tangent-line output therefore gives the estimate \(f(x)+f'(x)\,dx\).
The same reasoning follows from the definition of the derivative. For a nonzero increment \(h\), the actual change is \(f(x+h)-f(x)\). Differentiability at \(x\) means the quotient of that change by \(h\) approaches \(f'(x)\) as \(h\) approaches zero. Multiplying by \(h\) explains why the actual change is modeled, for a small increment, by \(f'(x)h\).
Here \(h\) is the input increment, so in an application we set \(h=dx\). This argument explains the estimate; it does not turn the approximation into an equality for an arbitrary finite increment.
Units in a Differential Estimate
Units provide a useful check on the calculation. If the input is measured in units \(U\) and the output in units \(V\), then \(f'(x)\) has units \(V/U\). Multiplying by \(dx\), which has units \(U\), gives \(dy\) in units \(V\), as required for an output change.
The input increment must also have the correct sign and units. For instance, if a measurement falls from \(x\) to \(x-0.2\), then \(dx=-0.2\), not \(0.2\). A negative differential predicts a decrease in the output; a positive differential predicts an increase.
Worked Examples
Worked Example: Estimating the Change in a Square Root
A square has area \(A=25\text{ cm}^2\). Its area increases by \(dA=0.4\text{ cm}^2\). Use \(s(A)=\sqrt{A}\) to estimate the change in its side length and the new side length.
First evaluate the side length at the starting area:
Differentiate with respect to area:
The derivative has units \(\text{cm}/\text{cm}^2\). Apply the differential formula:
Thus, the estimated increase in side length is \(0.04\text{ cm}\). Add it to the starting side length to get the estimated new side length:
Check the calculation by substituting the starting area and its derivative into \(s(25)+s'(25)dA\): \(5+(1/10)(0.4)=5.04\). The units of the final value are centimetres. For comparison, the exact new side is \(\sqrt{25.4}\text{ cm}\), approximately \(5.03984\text{ cm}\) to five decimal places; the differential result is an estimate, not an exact equality.
Worked Example: Estimating a Decrease in a Cubic Output
Let \(V(x)=x^3\), where \(x\) is a length measured in centimetres, so \(V\) is measured in cubic centimetres. If \(x\) decreases from \(2\text{ cm}\) to \(1.97\text{ cm}\), estimate the change in \(V\) and the new value.
The signed input increment is the new length minus the old length:
At the starting input,
Differentiate and evaluate the derivative at \(x=2\):
Now calculate the differential:
The estimated change is a decrease of \(0.36\text{ cm}^3\). The estimated new value is the starting value plus this signed change:
Check: \(V(2)+V'(2)dx=8+12(-0.03)=8-0.36=7.64\), in cubic centimetres. Direct evaluation gives \(V(1.97)=(1.97)^3=7.645373\text{ cm}^3\), so the actual change is \(-0.354627\text{ cm}^3\). The differential’s negative sign correctly predicts a decrease.
Worked Example: Estimating a Trigonometric Output
Use \(f(\theta)=\sin\theta\), with \(\theta\) measured in radians. Estimate the new value when \(\theta\) increases from \(\pi/6\) by \(0.02\) radians.
The starting output and the derivative at the starting angle are
Set \(d\theta=0.02\) radians. The differential estimate is
The sine output is dimensionless, so \(dy\) is dimensionless as well. Add the estimated change to the starting output:
Check by evaluating the tangent-line estimate directly: \(f(\pi/6)+f'(\pi/6)d\theta=0.5+(\sqrt{3}/2)(0.02)\approx0.517321\), rounded to six decimal places. The estimated new sine value is therefore \(0.517321\).
A Common Mistake: Confusing the Change With the New Value
Suppose \(f(x)=x^2\), the starting input is \(x=10\), and the new input is \(9.8\). The input increment is \(dx=9.8-10=-0.2\). Since \(f'(x)=2x\), the derivative at the starting input is \(f'(10)=20\). Therefore, \(dy=20(-0.2)=-4\). This is an estimated change, not the estimated new function value. The new value is \(f(10)+dy=100-4=96\).
A wrong calculation would report \(-4\) as the new value, omitting the original output. Another wrong approach would use \(9.8\) as \(dx\), even though \(9.8\) is the new input, not the increment. The correction is to subtract first: \(dx=9.8-10=-0.2\). As a check, the exact new value is \(f(9.8)=(9.8)^2=96.04\), close to the estimated value \(96\).
Using the Estimate in Applications
Differentials are useful when an application asks how a function’s output is expected to respond to a small change in its input. A calculation might predict an updated length, volume, cost, or other modeled quantity. In each case, the method is the same:
Subtract the original input from the new input: \(dx=x_{\text{new}}-x\).
Use \(f'(x)\), not the derivative evaluated at the new input.
Calculate \(dy=f'(x)\,dx\), retaining its sign and units.
Calculate \(f(x)+dy\) and state the result with output units, if the quantities have units.
These steps make clear what the result means. The differential estimates a change; adding it to the starting output gives the estimate at the new input. This distinction is especially important when an application asks for a new value rather than for the amount of change.
Check Your Understanding
For each question, identify whether the requested quantity is an estimated change or an estimated new value. Keep track of signs and units.
- For \(f(x)=\sqrt{x}\), the input increases from \(x=16\) to \(x=16.2\). Find \(dx\), \(dy\), and the estimated new function value.
- For \(g(t)=t^2\), the input decreases from \(t=3\) to \(t=2.9\). What is the differential estimate of the output change?
- If \(f'(x)\) has units of dollars per hour and \(dx\) is measured in hours, what units should \(dy\) have?
- In the differential formula, should the derivative be evaluated at the starting input or the new input? Explain briefly.
- A calculation gives \(dy=0.6\) and the original function value is \(f(x)=4.2\). What is the estimated new function value?
