Derivative Rules: Choose the First Step

Nablio · 10 min · core guidePublished October 2, 2026

The time estimate covers the core guide. Allow extra time for the optional practice.

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This three-question diagnostic is for students who have already seen the power, product, quotient and chain rules, plus derivatives of exponential and logarithmic functions. Start with a guided choice, diagnose an error, then choose the rules independently.

For each original question, identify the outermost operation: the last operation used to build the function from its parts. Choose the first derivative rule from that structure, then work inward. Here $x$ is a real input; a prime, as in $f^{\prime}(x)$ , means differentiation with respect to $x$ . Try each question before opening its answer.

1. Guided: a function inside a power

Begin with one inner expression and one outer power. The prompt gives two possible starting operations.

Try it yourself

Let $f(x)=(2x^2+3)^4$ . Is the outermost operation addition or raising to the fourth power? Name the first derivative rule, find $f^{\prime}(x)$ without expanding, and state the domain.

Show answer and reasoning

Read the outside first
The addition is inside the parentheses. The last operation raises the whole inner expression to the fourth power, so start with the chain rule. Set $u(x)=2x^2+3$ ; the outer function takes $u$ to $u^4$ .
Multiply the two rates
The outer derivative is $4u^3$ , and the inner derivative is $u^{\prime}(x)=4x$ . The chain rule multiplies them:
$\begin{aligned}f^{\prime}(x)&=4(2x^2+3)^3(4x)\\&=16x(2x^2+3)^3.\end{aligned}$
The factor $4x$ accounts for how the inner expression changes with $x$ ; keeping only $4(2x^2+3)^3$ would omit that change.
Keep the domain
Both the function and its derivative are polynomials, defined for every real $x$ . No denominator, square root or logarithm introduces a restriction.

2. Diagnose: multiplying derivatives

Now distinguish the rule for the whole function from the rule needed inside one factor. Explain the error before correcting it.

Try it yourself

Let $g(x)=(x-1)e^{3x}$ , where $e$ is the base of the natural exponential. A student writes $g^{\prime}(x)=(1)(3e^{3x})=3e^{3x}$ . Explain the rule-selection error, correct the derivative, and state its domain.

Show answer and reasoning

Identify the incorrect step
The outermost operation multiplies two functions of $x$ . The student differentiated each factor correctly but then multiplied those derivatives. A product needs the product rule, which adds two contributions. For $u(x)=x-1$ and $v(x)=e^{3x}$ ,
$(uv)^{\prime}=u^{\prime}v+uv^{\prime}.$
Use the chain rule inside the product
Here $u^{\prime}(x)=1$ . For the exponential, the inner exponent $3x$ has derivative $3$ , so the chain rule gives $v^{\prime}(x)=3e^{3x}$ . Therefore
$\begin{aligned}g^{\prime}(x)&=e^{3x}+3(x-1)e^{3x}\\&=(3x-2)e^{3x}.\end{aligned}$
The first term changes the linear factor while keeping the exponential; the second changes the exponential while keeping the linear factor. Both are needed.
Compare the answers and check the domain
At $x=0$ , the corrected derivative gives $g^{\prime}(0)=-2$ , whereas the student’s expression gives $3$ . Both original factors are defined and differentiable for every real $x$ , so that is also the derivative’s domain.

3. Independent: a logarithm in a quotient

Bring the same structure-first approach to a new expression. The solution stays closed while you choose the rules and track both input restrictions.

Try it yourself

Let $h(x)=\dfrac{\ln(3x+1)}{x-2}$ , where $\ln$ is the natural logarithm. State the real domain, explain which derivative rule comes first and which additional rule is needed, then find $h^{\prime}(x)$ . Leave the result in a form that shows your rule choices.

Show answer and reasoning

Restrict the inputs before differentiating
The logarithm requires $3x+1>0$ , so $x>-1/3$ . The denominator also requires $x\ne2$ . The function’s domain is therefore $(-1/3,2)\cup(2,\infty)$ ; the symbol $\cup$ joins the two intervals.
Choose the outer rule, then the inner rule
The outermost operation divides two functions, so the quotient rule is a direct starting point. Set $u(x)=\ln(3x+1)$ and $v(x)=x-2$ . The chain rule in the numerator gives $u^{\prime}(x)=3/(3x+1)$ , while $v^{\prime}(x)=1$ . Apply
$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime}v-uv^{\prime}}{v^2}.$
The numerator’s derivative is multiplied by the original denominator; then subtract the original numerator times the denominator’s derivative. The denominator is squared.
Substitute and retain the restrictions
$\begin{gathered}h^{\prime}(x)\\=\frac{\dfrac{3(x-2)}{3x+1}-\ln(3x+1)}{(x-2)^2}.\end{gathered}$
This derivative is valid on the same two open intervals as $h$ . Neither $x=-1/3$ nor $x=2$ may be restored by rearranging the formula. Rewriting the original quotient as $\ln(3x+1)(x-2)^{-1}$ is also valid on this domain; it uses the product and chain rules and gives the same derivative.

Optional extension: when the base and exponent both change

To work through this example, you’ll need the product rule, the chain rule and the derivatives of exponential and logarithmic functions. You’ll learn how taking a logarithm helps you differentiate a power when both its base and exponent change, then try a similar question yourself. This optional example goes beyond the three questions above and is not part of the Concordia MATH 203 early-midterm practice path.

In this section, $x$ is a real, dimensionless input, a prime means differentiation with respect to $x$ , $e$ is the natural exponential base, and $\ln$ is the natural logarithm.

Look at what stays fixed before choosing a rule. In $x^3$ , the exponent is the constant 3, so the power rule gives $3x^2$ . In $3^x$ , the base is the constant 3, so the exponential rule gives $3^x\ln3$ . In $x^x$ , both change. Neither of those constant-part formulas can be used unchanged.

Logarithmic differentiation makes the changing exponent into a factor. For a positive differentiable function $y=y(x)$ , the chain rule gives

$\frac{d}{dx}\ln y=\frac{y'}{y}.$

The numerator is $y'$ , not 1, because $y$ itself depends on $x$ .

Worked example: differentiate a variable power

Let $y=x^x$ on $x>0$ . This positive-base real power means $e^{x\ln x}$ , so $y>0$ and taking its logarithm is valid:

$\ln y=x\ln x.$

Differentiate both sides. The left side needs the chain rule; the right side needs the product rule because both factors change:

$\begin{aligned}\frac{y'}{y}&=1\cdot\ln x+x\frac1x\\&=\ln x+1.\end{aligned}$

Multiply by the original function to recover its derivative:

$\begin{gathered}y'=x^x(\ln x+1),\\x>0.\end{gathered}$

For example, at $x=1$ , the function value is 1 and its slope is 1. Writing the same function as $e^{x\ln x}$ and applying the chain rule gives the same derivative. The domain restriction is part of the argument: this derivation makes no claim about negative bases or the value at zero.

A tempting incorrect method

A student writes $(x^x)'=x\,x^{x-1}=x^x$ . That applies a constant-exponent power rule while the exponent is changing. It misses the contribution from that change. At $x=e$ , the incorrect expression gives $e^e$ , while the derivative above gives $2e^e$ . A familiar-looking power is not enough; first decide which parts depend on $x$ .

The general rule for a changing base and exponent

More generally, if $u$ and $v$ are differentiable on an open interval and $u(x)>0$ there, define $y=u(x)^{v(x)}=e^{v(x)\ln u(x)}$ . The same reasoning gives

$\begin{aligned}\frac{y'}y&=v'\ln u\\&\quad+v\frac{u'}u,\end{aligned}$

with all functions evaluated at $x$ . The first term tracks the changing exponent; the second tracks the changing base. Multiply by $y$ at the end.

Try a new base and exponent

Let $g(x)=(x+1)^{2x}$ on $x>-1$ . Find $g'(x)$ and $g'(0)$ . Explain why taking a logarithm is valid, and identify the contributions from the changing base and exponent before combining them.

Show hint

Set $y=g(x)$ . Take the logarithm first, then differentiate $2x\ln(x+1)$ as a product. Remember the derivative of $\ln y$ .

Show explained solution

On $x>-1$ , the base $x+1$ is positive and the real power is positive, so $\ln y=2x\ln(x+1)$ is valid. Differentiating gives

$\begin{aligned}\frac{y'}y&=2\ln(x+1)\\&\quad+\frac{2x}{x+1}.\end{aligned}$

The first term comes from differentiating the exponent $2x$ ; the second comes from differentiating the logarithm of the base. Therefore

$\begin{aligned}g'(x)&=(x+1)^{2x}\\&\quad\times\Bigl(2\ln(x+1)\\&\qquad+\frac{2x}{x+1}\Bigr).\end{aligned}$

At $x=0$ , $g(0)=1$ and both terms in parentheses are zero, so $g'(0)=0$ . This is an interior point of the stated domain. As a check, rewrite $g(x)=e^{2x\ln(x+1)}$ and apply the chain rule; it gives the same two contributions.

If the first decision is still unclear, bring either example or a basic question about powers and exponentials to a tutor. You can identify which parts change, work through the logarithm step together, then try a new function with guidance.

Tutoring sessions are currently offered in English.

Your next revision session

For question 1, check whether you included the inner derivative. For question 2, check whether both product-rule terms are present. For question 3, check the quotient-rule subtraction and both domain restrictions.
Mark the first step you missed: reading the structure, differentiating an inner function, applying the outer rule or keeping the domain. Redo that question with its answer closed.
If algebra or input restrictions are the obstacle, use the algebra and domain guides linked below. Then choose another derivative question from your own course materials and explain your first rule before calculating.

Dawson course context and further practice

These three core questions, written independently by Nablio, practise combined derivative rules relevant to Dawson 201-SN2-RE. Questions 6 and 10 of the Winter 2025 final provide topic-level context; the functions and prompts here were written independently. Use your section’s outline for assessed topics.

Useful for

Work on the step you’re missing.

Studying differential calculus at CEGEP? See CEGEP calculus tutoring to find your college’s course page. For individual help, bring the first derivative step you could not explain.

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Derivative Rules: Choose the First Step

1. Guided: a function inside a power

Try it yourself

Read the outside first

Multiply the two rates

Keep the domain

2. Diagnose: multiplying derivatives

Try it yourself

Identify the incorrect step

Use the chain rule inside the product

Compare the answers and check the domain

3. Independent: a logarithm in a quotient

Try it yourself

Restrict the inputs before differentiating

Choose the outer rule, then the inner rule

Substitute and retain the restrictions

Optional extension: when the base and exponent both change

Worked example: differentiate a variable power

A tempting incorrect method

The general rule for a changing base and exponent

Try a new base and exponent

Your next revision session

Dawson course context and further practice

Useful for

Work on the step you’re missing.

Free study guides and practice

Before Calculus: 10 Algebra Skills You Should Know

Cancelling Factors: Why the Missing Input Stays Missing

Sketch a rational function without crossing its domain breaks