FTC Part 1 with Chain Rule

4 min read

Reference: Stewart §5.3

Textbook Reference


Primary source	OpenStax Calculus Volume 2, Section 1.3: “The Fundamental Theorem of Calculus”
Direct link	https://openstax.org/books/calculus-volume-2/pages/1-3-the-fundamental-theorem-of-calculus
Original home	OpenStax Calculus Volume 1, Section 5.3: “The Fundamental Theorem of Calculus”
Original link	https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus
Stewart	Calculus (Stewart), Section 5.3: “The Fundamental Theorem of Calculus”

This lesson combines FTC Part 1 (from the section above) with the chain rule from differential calculus. The chain rule is in OpenStax Calculus Volume 1, Section 3.6 (https://openstax.org/books/calculus-volume-1/pages/3-6-the-chain-rule). Both sources are free and openly licensed.

When the Upper Limit Is Not Just $x$

In the basic FTC Part 1, the upper limit of integration is simply $x$. But what if the upper limit is something more complicated, like $x^2$ or $\sin x$? You are no longer differentiating with respect to the upper limit directly: you need the chain rule.

This is like reading a speedometer while driving: if the speedometer reading is $v(t)$, and you want to know how fast the reading changes as you accelerate, you need to consider both the speed itself AND how quickly you are changing it. It is a rate of a rate.

Prerequisite Map

Quick Reference

Property	Value
Concept	Fundamental Theorem of Calculus
Course	MATH161 (Calculus I)
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

The Formula

If $g(x) = \displaystyle\int_a^{u(x)} f(t)\, dt$, where $u(x)$ is a differentiable function, then:

$$\boxed{g'(x) = f(u(x)) \cdot u'(x)}$$

In Leibniz notation:

$$\frac{d}{dx}\int_a^{u(x)} f(t)\, dt = f(u(x)) \cdot \frac{du}{dx}$$

Why This Works

Think of it as a composition of functions:

First, FTC1 gives us a function $G(u) = \int_a^u f(t)\, dt$ with $G'(u) = f(u)$
Then we are computing $g(x) = G(u(x))$
By the chain rule: $g'(x) = G'(u(x)) \cdot u'(x) = f(u(x)) \cdot u'(x)$

Visualization

                    ┌──────────────┐
         x  ────►   │   u = u(x)   │  ────►  u
                    └──────────────┘
                           │
                           ▼
                    ┌──────────────┐
         u  ────►   │ G(u) = ∫f dt │  ────►  G
                    └──────────────┘

    Derivative = f(u(x)) · u'(x)
                    ▲          ▲
                    │          │
              from FTC1   from chain rule

The Pattern to Remember

When differentiating $\displaystyle\int_a^{u(x)} f(t)\, dt$:

Substitute the upper limit into $f$: write $f(u(x))$
Multiply by the derivative of the upper limit: $\times\, u'(x)$

Both Limits Variable

For the general case $\displaystyle\int_{v(x)}^{u(x)} f(t)\, dt$, split it:

$$\int_{v(x)}^{u(x)} f(t)\, dt = \int_a^{u(x)} f(t)\, dt - \int_a^{v(x)} f(t)\, dt$$

Then differentiate each piece:

$$\frac{d}{dx}\int_{v(x)}^{u(x)} f(t)\, dt = f(u(x)) \cdot u'(x) - f(v(x)) \cdot v'(x)$$

Practice Problems

Level 1 Simple Quadratic Upper Limit

Find $\displaystyle\frac{d}{dx}\int_1^{x^2} \cos(t)\, dt$.

Thought Process

Identify the pieces:

Lower limit: 1 (constant, good)
Upper limit: $u(x) = x^2$
Integrand: $f(t) = \cos(t)$

Apply the formula: $f(u(x)) \cdot u'(x)$

$f(u(x)) = \cos(x^2)$
$u'(x) = 2x$

Multiply them together.

Show Answer

Let $u = x^2$. By FTC1 with chain rule:

$$\frac{d}{dx}\int_1^{x^2} \cos(t)\, dt = \cos(x^2) \cdot 2x = 2x\cos(x^2)$$

Level 2 Trigonometric Upper Limit

Find $h'(x)$ if $h(x) = \displaystyle\int_0^{\sin x} e^{t^2}\, dt$.

Thought Process

The upper limit is $u(x) = \sin x$, so $u'(x) = \cos x$.

The integrand is $f(t) = e^{t^2}$.

At the upper limit: $f(u(x)) = f(\sin x) = e^{(\sin x)^2} = e^{\sin^2 x}$.

Apply the formula: multiply $f(u(x))$ by $u'(x)$.

Show Answer

With $u(x) = \sin x$ and $f(t) = e^{t^2}$:

$$h'(x) = e^{(\sin x)^2} \cdot \cos x = e^{\sin^2 x} \cos x$$

Level 3 Variable in Lower Limit

Find $\displaystyle\frac{d}{dx}\int_{x^3}^{1} \frac{1}{1+t^4}\, dt$.

Thought Process

The variable $x$ is in the lower limit, not the upper. First, flip the integral:

$$\int_{x^3}^{1} f(t)\, dt = -\int_1^{x^3} f(t)\, dt$$

Now the upper limit is $u(x) = x^3$ and we can apply FTC1 with chain rule.

Do not forget the negative sign from flipping.

Show Answer

First, reverse the limits:

$$\int_{x^3}^{1} \frac{1}{1+t^4}\, dt = -\int_1^{x^3} \frac{1}{1+t^4}\, dt$$

Now apply FTC1 with chain rule (and the negative sign):

$$\frac{d}{dx}\int_{x^3}^{1} \frac{1}{1+t^4}\, dt = -\frac{1}{1+(x^3)^4} \cdot 3x^2 = -\frac{3x^2}{1+x^{12}}$$

Level 4 Both Limits Variable

Differentiate $g(x) = \displaystyle\int_{\sqrt{x}}^{x^2} \ln(1 + t)\, dt$.

Thought Process

Both limits are functions of $x$. Use the splitting technique:

$$\int_{\sqrt{x}}^{x^2} f(t)\, dt = \int_a^{x^2} f(t)\, dt - \int_a^{\sqrt{x}} f(t)\, dt$$

For some constant $a$ (the choice does not matter since it cancels).

Differentiate each piece using FTC1 + chain rule:

First integral: upper limit $u = x^2$, so derivative is $f(x^2) \cdot 2x$
Second integral: upper limit $v = \sqrt{x} = x^{1/2}$, so derivative is $f(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$

Show Answer

Split using an intermediate constant:

$$g(x) = \int_1^{x^2} \ln(1+t)\, dt - \int_1^{\sqrt{x}} \ln(1+t)\, dt$$

Differentiate each term:

$$g'(x) = \ln(1 + x^2) \cdot 2x - \ln(1 + \sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$$

Simplify:

$$g'(x) = 2x\ln(1 + x^2) - \frac{\ln(1 + \sqrt{x})}{2\sqrt{x}}$$

Level 5 Tangent Line Problem

Let $F(x) = \displaystyle\int_\pi^{x^2} \frac{\cos t}{t}\, dt$.

Find the equation of the tangent line to the curve $y = F(x)$ at the point where $x = \sqrt{\pi}$.

Thought Process

For the tangent line $y - y_0 = m(x - x_0)$, we need:

The point $(x_0, y_0) = (\sqrt{\pi}, F(\sqrt{\pi}))$
The slope $m = F'(\sqrt{\pi})$

For the $y$-coordinate: When $x = \sqrt{\pi}$, the upper limit is $(\sqrt{\pi})^2 = \pi$, so $$F(\sqrt{\pi}) = \int_\pi^\pi \frac{\cos t}{t}\, dt = 0$$

For the slope: Use FTC1 + chain rule to find $F'(x)$, then evaluate at $x = \sqrt{\pi}$.

Show Answer

Step 1: Find the point.

At $x = \sqrt{\pi}$: $$F(\sqrt{\pi}) = \int_\pi^\pi \frac{\cos t}{t}\, dt = 0$$

So the point is $(\sqrt{\pi}, 0)$.

Step 2: Find the slope.

Using FTC1 with chain rule (upper limit is $u = x^2$):

$$F'(x) = \frac{\cos(x^2)}{x^2} \cdot 2x = \frac{2\cos(x^2)}{x}$$

At $x = \sqrt{\pi}$: $$F'(\sqrt{\pi}) = \frac{2\cos(\pi)}{\sqrt{\pi}} = \frac{2(-1)}{\sqrt{\pi}} = -\frac{2}{\sqrt{\pi}}$$

Step 3: Write the tangent line.

$$y - 0 = -\frac{2}{\sqrt{\pi}}(x - \sqrt{\pi})$$

$$y = -\frac{2}{\sqrt{\pi}}x + 2$$

Or equivalently: $y = -\frac{2\sqrt{\pi}}{\pi}x + 2$

Common Misconceptions

Common misconception

differentiating $\int_a^{u(x)} f(t)\,dt$ gives $f(u(x))$ without multiplying by $u'(x)$.

This is the composition-is-not-chaining error. The accumulation function $G(s) = \int_a^s f(t)\,dt$ satisfies $G'(s) = f(s)$ by FTC Part 1. Differentiating $G(u(x))$ with respect to $x$ requires the chain rule: $\frac{d}{dx}G(u(x)) = G'(u(x))\cdot u'(x) = f(u(x))\cdot u'(x)$. For example, $\frac{d}{dx}\int_1^{x^4}\sec t\,dt = \sec(x^4)\cdot 4x^3$, not merely $\sec(x^4)$. Omitting $u'(x)$ is the same oversight as writing $\frac{d}{dx}\sin(x^4) = \cos(x^4)$ instead of $4x^3\cos(x^4)$.

Mastery Checklist

I can differentiate $\int_a^{u(x)} f(t)\, dt$ when the upper limit is a function of $x$
I remember to multiply by $u'(x)$ (the chain rule factor)
I can handle the case when $x$ appears in the lower limit (flip and negate)
I can differentiate when both limits are functions of $x$
I understand why this is a composition that requires the chain rule

Mental Model

The Odometer Analogy: Imagine $\int_a^{u(x)} f(t)\, dt$ as reading an odometer. If $u(x)$ tells you “how far along the road you are” and $f$ is “the scenery at each point,” then:

FTC1 says the scenery changes as you move
The chain rule accounts for how fast you are moving along the road

So the total rate of scenery change = (scenery at current position) × (speed of travel).

Common Mistakes

Mistake	Correction
Forgetting the $u'(x)$ factor	Always multiply by the derivative of the upper limit
Sign error when lower limit has $x$	Flip limits first, then remember the negative sign
Evaluating $f$ at $x$ instead of $u(x)$	Substitute the entire upper limit into $f$

Connections

Looking back:

Basic FTC Part 1 is the special case where $u(x) = x$
This is a direct application of the chain rule from differentiation

Looking ahead:

FTC Part 2 shows how to evaluate definite integrals using antiderivatives
The Leibniz integral rule (later courses) handles even more general cases

Real-world connections:

In physics: when displacement depends on a function of time in a complicated way
In economics: when cost depends on production level, which itself depends on time

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FTC Part 1	Skills Index	FTC Part 2

Last updated: 2026-01-22