Computing Differentials From Function Formulas

From a Function Formula to Its Differential

A differential calculation starts with a function $y=f(x)$, a point at which to evaluate the derivative, and an input differential $dx$. The output differential is obtained by multiplying that input differential by the derivative at the specified point. The calculation is short, but keeping the evaluation point visible prevents a common source of mistakes.

This tutorial applies that procedure to polynomial, rational, trigonometric, and exponential functions. The goal is to calculate the differential reliably, not to treat $dx$ as an ordinary number to be silently dropped or to leave the derivative unevaluated.

Definition: Let $f$ be differentiable at $x=a$. The differential of $y=f(x)$, evaluated at $x=a$, is $$ \left.dy\right|_{x=a}=f'(a)\,dx. $$ Here $dx$ is the input differential, and $f'(a)$ is the derivative evaluated at the stated point. The result is a linear expression in $dx$.

If the calculation is being made at a variable point $x$, the same rule is written $dy=f'(x)\,dx$. If a particular point is given, write $f'(a)$, or otherwise clearly indicate that the derivative has been evaluated there. A differential calculation requires differentiability at the evaluation point. For rational functions, the point must also lie in the function’s domain.

A Reliable Calculation Sequence

Use the same sequence for each function. First differentiate the formula with respect to its input. Next substitute the specified point into the derivative and simplify the resulting coefficient. Finally multiply that coefficient by $dx$. If a numerical value of $dx$ is supplied, substitute it after finding the coefficient.

Differentiate.
Find $f'(x)$ using a differentiation rule that applies to the given formula.

Evaluate.
Substitute the stated input value $a$ into $f'(x)$ and simplify.

Multiply by $dx$.
Write the evaluated derivative coefficient times the input differential, keeping any supplied units.

Check.
Confirm that the coefficient in the result equals the derivative of the original function evaluated at the stated point.

The final check is especially useful when a formula involves a quotient, a composite function, or several terms. It makes the role of the evaluation point explicit rather than relying on an unevaluated derivative.

Polynomial Differentials

For a polynomial, differentiate each term using the power rule and the constant-multiple rule. The power rule states that the derivative of $x^n$ is $nx^{n-1}$ for a nonnegative integer $n$. Constants have derivative zero. After differentiating, evaluate the resulting polynomial at the requested point.

Worked Example: A Polynomial at a Specified Point

Let $y=3x^4-2x^3+5x-7$. Find the differential at $x=2$ when $dx=-0.03$.

Differentiate each term:

\frac{dy}{dx}=f'(x)=12x^3-6x^2+5.

Evaluate the derivative at $x=2$:

$$ f'(2)=12(2^3)-6(2^2)+5 =12(8)-6(4)+5 =96-24+5 =77. $$

Therefore the differential at the specified point is $77dx$. Substituting $dx=-0.03$ gives

\left.dy\right|_{x=2}=77(-0.03)=-2.31.

Check: differentiating the original polynomial gives $12x^3-6x^2+5$, whose value at $2$ is $77$. Thus the coefficient in the calculated differential is $f'(2)$, as required.

Rational Differentials

A rational function is a quotient of polynomials. To use the quotient rule, write the numerator as $u(x)$ and the denominator as $v(x)$. At points where $v(x)\ne 0$, the derivative is

\left(\frac{u}{v}\right)'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}.

The squared denominator matters. A rational function and its differential are only defined at points where the original denominator is nonzero; a derivative formula does not make a point outside the function’s domain valid.

Worked Example: A Quotient with a Nonzero Denominator

Let $y=\dfrac{2x+1}{x-1}$. Find the differential at $x=3$ when $dx=-0.08$.

The denominator $x-1$ is nonzero at $x=3$, so the function is defined there. Set $u(x)=2x+1$ and $v(x)=x-1$. Then $u'(x)=2$ and $v'(x)=1$. Apply the quotient rule:

f'(x)=\frac{2(x-1)-(2x+1)(1)}{(x-1)^2} =\frac{2x-2-2x-1}{(x-1)^2} =\frac{-3}{(x-1)^2}.

Evaluate at $x=3$:

f'(3)=\frac{-3}{(3-1)^2} =\frac{-3}{2^2} =-\frac{3}{4}.

Multiply by the specified input differential:

\left.dy\right|_{x=3}=-\frac{3}{4}dx =-\frac{3}{4}(-0.08) =0.06.

Check: substituting $x=3$ into the quotient-rule derivative gives $-3/4$, which is exactly the coefficient of $dx$ in the result.

Trigonometric Differentials

For trigonometric functions, use the usual derivative formulas with angles measured in radians. For example, $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$. The evaluation point must be substituted into these derivative formulas, not into the original function in place of differentiation.

Worked Example: A Sum of Sine and Cosine

Let $y=\sin x+2\cos x$. Find the differential at $x=\pi/3$ when $dx=0.02$ radians.

Differentiate each term:

f'(x)=\cos x-2\sin x.

Use $\cos(\pi/3)=1/2$ and $\sin(\pi/3)=\sqrt{3}/2$:

f'\left(\frac{\pi}{3}\right) =\cos\left(\frac{\pi}{3}\right)-2\sin\left(\frac{\pi}{3}\right) =\frac{1}{2}-2\left(\frac{\sqrt{3}}{2}\right) =\frac{1}{2}-\sqrt{3}.

Thus the differential, with the evaluation point and input differential retained, is

\left.dy\right|_{x=\pi/3} =\left(\frac{1}{2}-\sqrt{3}\right)dx =\left(\frac{1}{2}-\sqrt{3}\right)(0.02) =0.01-0.02\sqrt{3}.

Check: the derivative of the original sum is $\cos x-2\sin x$; at $\pi/3$ its value is $1/2-\sqrt{3}$. This matches the coefficient multiplying $dx$. The input differential is in radians, as required for these derivative formulas.

Exponential Differentials

The derivative of $e^x$ is $e^x$. More generally, for a positive constant $b$, the derivative of $b^x$ is $b^x\ln b$, where $\ln$ denotes the natural logarithm. If the exponent is a function of $x$, the chain rule supplies the derivative of that exponent as an additional factor.

Worked Example: A Composite Exponential

Let $y=e^{2x}$. Find the differential at $x=0$ when $dx=0.05$.

The exponent is $2x$, whose derivative is $2$. By the chain rule,

f'(x)=e^{2x}\cdot 2=2e^{2x}.

Evaluate at $x=0$:

f'(0)=2e^{2(0)}=2e^0=2(1)=2.

Therefore,

\left.dy\right|_{x=0}=2dx=2(0.05)=0.10.

Check: differentiating $e^{2x}$ gives $2e^{2x}$, and this derivative equals $2$ at zero. The computed differential therefore has the correct coefficient, $f'(0)=2$.

Why the Differential Has This Form

The formula $dy=f'(a)\,dx$ comes from the linear model of $f$ at $a$. For a function differentiable at $a$, its linearization at that point is the function

$$ L(x)=f(a)+f'(a)(x-a). $$

This line has the same value and slope as $f$ at the base point $a$. If the input displacement in this linear model is denoted by $dx$, the output displacement of the line is its slope multiplied by its input displacement. Indeed, putting $x=a+dx$ into $L$ gives

L(a+dx)-L(a) =\bigl[f(a)+f'(a)dx\bigr]-f(a) =f'(a)dx.

Thus the differential is precisely the output displacement assigned by the tangent-line model. This derivation explains both why the derivative is evaluated at the base point and why the result is multiplied by the input differential. It also shows why the point cannot be omitted when a numerical evaluation is requested.

Key takeaway: To compute a differential at $x=a$, calculate $f'(a)$ first and then multiply by $dx$. The result is $\left.dy\right|_{x=a}=f'(a)\,dx$.

A Common Mistake: Leaving the Derivative Unevaluated

A frequent error is to differentiate correctly but stop before evaluating at the specified point. For example, for $y=\sin x$, the derivative is $\cos x$. At $x=\pi/6$ with $dx=0.1$, writing $dy=\cos x\,dx$ has not yet completed the requested evaluation. Worse, replacing $\cos x$ by $\sin x$ confuses the original function with its derivative.

The correction is to substitute the given point into the derivative before multiplying by $dx$:

f'\left(\frac{\pi}{6}\right) =\cos\left(\frac{\pi}{6}\right) =\frac{\sqrt{3}}{2}, \qquad \left.dy\right|_{x=\pi/6} =\frac{\sqrt{3}}{2}(0.1) =\frac{\sqrt{3}}{20}.

Check: the derivative of $\sin x$ is $\cos x$, and its value at $\pi/6$ is $\sqrt{3}/2$. That evaluated value, rather than the unevaluated expression $\cos x$, is the coefficient of $dx$.

Checking the Scope of a Calculation

Before finalizing a differential, check three things: the derivative rule matches the function, the evaluation point is allowed by the function’s domain, and the coefficient in the answer is the derivative at that point. For trigonometric formulas, also check that angles and input differentials use radians. For a rational function, inspect the denominator before evaluating.

These checks are useful because a differential is compact notation: one missing evaluation or one incorrect derivative factor changes the entire result. Keeping the structure visible—derivative, point, input differential—makes the calculation easy to audit.

Check Your Understanding

For each question, show the derivative, evaluate it at the stated point, and then multiply by the given $dx$.

For $y=2x^3-x$, find the differential at $x=-1$ when $dx=0.04$.
For $y=\dfrac{x+2}{x}$, find the differential at $x=2$ when $dx=-0.1$. Confirm that the evaluation point is in the domain.
For $y=\cos x$, find the differential at $x=\pi/3$ when $dx=0.02$ radians.
For $y=5^x$, find the differential at $x=0$ when $dx=0.1$. Leave the exact coefficient in terms of $\ln 5$.
In one sentence, explain why a differential calculation at $x=a$ uses $f'(a)$, rather than $f'(x)$, as the coefficient of $dx$.