Putting Rules Together: Multi-Step Differentiation and Simplification

A reliable workflow for multi-step derivatives

When expressions combine powers, products, quotients, and compositions, the main difficulty is not any single rule—it is choosing the right rule at the right time and keeping algebra organized. Use this consistent workflow:

1. Rewrite into a form that makes structure obvious (exponents instead of radicals, negative exponents instead of fractions when helpful, simplify inside first).
2. Choose rules by identifying the outermost operation: sum/difference, product, quotient, or a composition (something “inside” something).
3. Differentiate carefully with clear grouping symbols.
4. Simplify with a goal: combine like factors, reduce complex fractions, and factor when it helps interpretation or later steps.

Rewrite moves that prevent mistakes

• Radicals: \sqrt{x}=x^{1/2}, \sqrt[3]{x^2}=x^{2/3}.
• Fractions: \frac{1}{x^3}=x^{-3}. But keep a quotient if it clarifies “top vs bottom” structure.
• Complex fractions: \frac{\frac{x+1}{x}}{x^2}=\frac{x+1}{x^3}.
• Factor before differentiating only if it clearly simplifies the rule choice (e.g., cancel common factors in a quotient), but never cancel across addition (e.g., \frac{x^2+1}{x} cannot cancel an x).

Examples where the expressions look similar but the rule choice changes

Example 1: Same numerator, different denominators

A. Differentiate f(x)=\frac{x^2+1}{x^3}.

Step 1 (rewrite): f(x)=(x^2+1)x^{-3} (product form).

Step 2 (choose rules): Product rule on (x^2+1) and x^{-3}.

Step 3 (differentiate):

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f'(x)=(2x)\,x^{-3}+(x^2+1)(-3)x^{-4}

Step 4 (simplify): factor out the highest common power x^{-4}:

f'(x)=x^{-4}\big(2x\cdot x -3(x^2+1)\big)=x^{-4}(2x^2-3x^2-3)

f'(x)=x^{-4}(-x^2-3)= -\frac{x^2+3}{x^4}

B. Differentiate g(x)=\frac{x^2+1}{x+3}.

Step 1 (rewrite): already a clear quotient.

Step 2 (choose rules): Quotient rule (top and bottom are both nontrivial).

Step 3 (differentiate):

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

Step 4 (simplify): expand only what you need:

g'(x)=\frac{2x^2+6x-x^2-1}{(x+3)^2}=\frac{x^2+6x-1}{(x+3)^2}

Notice how \frac{x^2+1}{x^3} became easiest after rewriting as a product with a negative exponent, while \frac{x^2+1}{x+3} stays naturally a quotient.

Example 2: Product vs composition (looks like “something squared”)

A. Differentiate h(x)=(x^2+1)^2.

Rule choice: composition (outer square applied to inner x^2+1).

h'(x)=2(x^2+1)\cdot (2x)=4x(x^2+1)

B. Differentiate k(x)=(x^2+1)\,x^2.

Rule choice: product (two factors multiplied).

k'(x)=(2x)\,x^2+(x^2+1)(2x)=2x^3+2x(x^2+1)=4x^3+2x

They look similar, but one is a single “inside squared,” the other is multiplication.

Example 3: Quotient with a composite numerator

Differentiate m(x)=\frac{\sqrt{x^2+1}}{x}.

Step 1 (rewrite): m(x)=\frac{(x^2+1)^{1/2}}{x}.

Step 2 (choose rules): quotient rule; numerator also needs chain rule.

Step 3 (differentiate):

m'(x)=\frac{\left(\frac{1}{2}(x^2+1)^{-1/2}\cdot 2x\right)\cdot x-(x^2+1)^{1/2}\cdot 1}{x^2}

Step 4 (simplify): the numerator derivative simplifies to \frac{x}{\sqrt{x^2+1}}:

m'(x)=\frac{\left(\frac{x}{\sqrt{x^2+1}}\right)x-\sqrt{x^2+1}}{x^2}

Combine into one fraction in the top:

m'(x)=\frac{\frac{x^2-(x^2+1)}{\sqrt{x^2+1}}}{x^2}=\frac{-1}{x^2\sqrt{x^2+1}}

Checkpoint: simplification goals (choose the form that matches the task)

Goal 1: Simplify for quick evaluation at a point

If you need f'(2), a simplified expression reduces arithmetic and error risk.

Example: from above, m'(x)=-\frac{1}{x^2\sqrt{x^2+1}} makes m'(2)=-\frac{1}{4\sqrt{5}} immediate. If you left it as a large quotient-rule expression, evaluation would be slower and more error-prone.

Goal 2: Simplify for sign analysis (increasing/decreasing, positivity/negativity)

For sign, you want a product of factors and a clearly positive denominator.

Example: f'(x)=-\frac{x^2+3}{x^4} is ideal for sign analysis: x^4>0 for x\neq 0 and x^2+3>0 always, so f'(x)<0 for all x\neq 0.

Goal 3: Leave in factored form when it reveals zeros or cancellations

Factored form helps identify where the derivative is zero (critical points) or where expressions simplify later.

Example: if you get p'(x)=4x(x^2+1), keeping it factored instantly shows p'(x)=0 at x=0 (and no other real zeros since x^2+1>0).

Cumulative worked examples (full workflow)

Example 4: Product + chain + power

Differentiate y=x^2\,(3x-1)^5.

Rewrite: already a product of two factors.

Choose rules: product rule; second factor needs chain rule.

y'=(2x)(3x-1)^5 + x^2\cdot 5(3x-1)^4\cdot 3

Simplify (factor common pieces): common factor x(3x-1)^4:

y'=x(3x-1)^4\big(2(3x-1)+15x\big)

y'=x(3x-1)^4(6x-2+15x)=x(3x-1)^4(21x-2)

Example 5: Quotient where rewriting avoids quotient rule

Differentiate q(x)=\frac{x^2+1}{x^3} again, but this time simplify first.

Rewrite:

q(x)=\frac{x^2}{x^3}+\frac{1}{x^3}=x^{-1}+x^{-3}

Differentiate:

q'(x)=-x^{-2}-3x^{-4}= -\frac{1}{x^2}-\frac{3}{x^4}= -\frac{x^2+3}{x^4}

This matches the earlier result and shows an alternative workflow: algebraic simplification can reduce rule complexity.

Example 6: Nested composition inside a quotient

Differentiate r(x)=\frac{(2x+1)^3}{\sqrt{x}}.

Rewrite: r(x)=(2x+1)^3\,x^{-1/2} (product form).

Choose rules: product rule; first factor uses chain rule.

r'(x)=\big(3(2x+1)^2\cdot 2\big)x^{-1/2} + (2x+1)^3\left(-\frac{1}{2}x^{-3/2}\right)

Simplify: factor (2x+1)^2 x^{-3/2}:

r'(x)=(2x+1)^2 x^{-3/2}\left(6x - \frac{1}{2}(2x+1)\right)

r'(x)=(2x+1)^2 x^{-3/2}\left(6x - x - \frac{1}{2}\right)=(2x+1)^2 x^{-3/2}\left(5x-\frac{1}{2}\right)

r'(x)=\frac{(2x+1)^2(10x-1)}{2x^{3/2}}

Exercises (with fully worked solutions)

Exercise 1

Differentiate f(x)=\frac{(x^2+1)^4}{x}.

Solution: Rewrite as product: f(x)=(x^2+1)^4 x^{-1}.

f'(x)=\big(4(x^2+1)^3\cdot 2x\big)x^{-1} + (x^2+1)^4(-1)x^{-2}

Simplify:

f'(x)=8x(x^2+1)^3\cdot x^{-1}-(x^2+1)^4 x^{-2}

f'(x)=8(x^2+1)^3-\frac{(x^2+1)^4}{x^2}

Factor (x^2+1)^3 and combine:

f'(x)=(x^2+1)^3\left(8-\frac{x^2+1}{x^2}\right)=(x^2+1)^3\left(\frac{8x^2-(x^2+1)}{x^2}\right)

f'(x)=\frac{(x^2+1)^3(7x^2-1)}{x^2}

Exercise 2

Differentiate g(x)=\frac{x^2+1}{(x+3)^2}.

Solution: Treat as product with a negative exponent: g(x)=(x^2+1)(x+3)^{-2}.

g'(x)=(2x)(x+3)^{-2}+(x^2+1)(-2)(x+3)^{-3}

Factor (x+3)^{-3}:

g'(x)=(x+3)^{-3}\big(2x(x+3)-2(x^2+1)\big)

g'(x)=\frac{2x(x+3)-2(x^2+1)}{(x+3)^3}=\frac{2x^2+6x-2x^2-2}{(x+3)^3}

g'(x)=\frac{6x-2}{(x+3)^3}=\frac{2(3x-1)}{(x+3)^3}

Exercise 3

Differentiate h(x)=\sqrt{x}\,(x^2+1)^{-1}.

Solution: Rewrite: h(x)=x^{1/2}(x^2+1)^{-1} (product).

h'(x)=\left(\frac{1}{2}x^{-1/2}\right)(x^2+1)^{-1}+x^{1/2}\left(-(x^2+1)^{-2}\cdot 2x\right)

Simplify and factor:

h'(x)=\frac{1}{2}x^{-1/2}(x^2+1)^{-1}-2x^{3/2}(x^2+1)^{-2}

h'(x)=x^{-1/2}(x^2+1)^{-2}\left(\frac{1}{2}(x^2+1)-2x^2\right)

h'(x)=x^{-1/2}(x^2+1)^{-2}\left(\frac{1}{2}-\frac{3}{2}x^2\right)=\frac{1-3x^2}{2\sqrt{x}(x^2+1)^2}

Exercise 4

Differentiate p(x)=\frac{(x-1)(x+2)}{x^2}.

Solution: Simplify first by expanding the numerator or rewriting as powers. Expand numerator:

p(x)=\frac{x^2+x-2}{x^2}=1+\frac{1}{x}-\frac{2}{x^2}=1+x^{-1}-2x^{-2}

Differentiate term-by-term:

p'(x)=0- x^{-2}+4x^{-3}= -\frac{1}{x^2}+\frac{4}{x^3}=\frac{-x+4}{x^3}

Error spotting (diagnose and correct)

Error Spotting 1: Wrong denominator power in a quotient

Problem: y=\frac{x^2+1}{x+3}. A student writes:

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

What’s wrong: In the quotient rule, the denominator must be squared: (x+3)^2, not (x+3).

Correct derivative:

y'=\frac{2x(x+3)-(x^2+1)}{(x+3)^2}=\frac{x^2+6x-1}{(x+3)^2}

Error Spotting 2: Treating a product as a power

Problem: y=(x^2+1)x^2. A student writes:

y'=2(x^2+1)x

What’s wrong: That derivative corresponds to (x^2+1)^2 (a composition), not a product. The original function multiplies two factors, so product rule is required.

Correct derivative:

y'=(2x)x^2+(x^2+1)(2x)=4x^3+2x

Error Spotting 3: Losing the inner derivative in a chain

Problem: y=\sqrt{x^2+1}=(x^2+1)^{1/2}. A student writes:

y'=\frac{1}{2}(x^2+1)^{-1/2}

What’s wrong: The inner derivative of x^2+1 (which is 2x) is missing.

Correct derivative:

y'=\frac{1}{2}(x^2+1)^{-1/2}\cdot 2x=\frac{x}{\sqrt{x^2+1}}