Derivatives

Derivatives and Tangent Lines

Example

Find $f'(2)$ for $f(x) = x^2 - 3x + 1$ using the limit definition.

Example

Find $f'(x)$ for $f(x) = \frac{1}{x}$ using the limit definition.

Example

Find $f'(x)$ for $f(x) = \sqrt{x}$ using the limit definition.

Multiply numerator and denominator by $\sqrt{x+h} + \sqrt{x}$:

This confirms the power rule for $n = \frac{1}{2}$: $\frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. $\blacksquare$

Theorem. If $f$ is differentiable at $a$Then $f$ is continuous at $a$.

Proof. We need to show $\displaystyle\lim_{x \to a} f(x) = f(a)$. Consider:

Therefore, $\displaystyle\lim_{x \to a} f(x) = f(a)$So $f$ is continuous at $a$. $\blacksquare$

The converse is false: $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there Because the left-hand and right-hand derivatives disagree.

A function fails to be differentiable at a point if:

1. There is a corner (left and right derivatives differ), e.g., $|x|$ at $x = 0$.
2. There is a cusp, e.g., $f(x) = x^{2/3}$ at $x = 0$.
3. There is a vertical tangent, e.g., $f(x) = \sqrt[3]{x}$ at $x = 0$.
4. There is a discontinuity.

Proof that $|x|$ is not differentiable at $x = 0$.

Left-hand derivative: $\displaystyle\lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$.

Right-hand derivative: $\displaystyle\lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.

Since $-1 \neq 1$The derivative does not exist. $\blacksquare$

Proof that $f(x) = x^{2/3}$ has a cusp at $x = 0$.

As x \to 0^+$$f'(x) \to +\infty. As x \to 0^-$$f'(x) \to -\infty.

The left and right derivatives are both infinite but with opposite signs, creating a cusp. The Tangent line approaches vertical from both sides. $\blacksquare$

Let $c$ be a constant and $n$ a real number:

1. Constant rule: $\displaystyle\frac{d}{dx}[c] = 0$
2. Power rule: $\displaystyle\frac{d}{dx}[x^n] = nx^{n-1}$
3. Constant multiple rule: $\displaystyle\frac{d}{dx}[cf(x)] = cf'(x)$
4. Sum/Difference rule: $\displaystyle\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$

Proof sketch. Write $F(x) = f(x)g(x)$ and expand the difference quotient $\frac{F(x+h) - F(x)}{h}$ by adding and subtracting $f(x+h)g(x)$:

Taking limits as $h \to 0$ gives the result. $\blacksquare$

Proof. Let $F(x) = \frac{f(x)}{g(x)}$So $f(x) = F(x)g(x)$. Differentiating using the product Rule:

$\blacksquare$

Mnemonic: “Low d-High minus High d-Low, draw a line and square below.”

Why the chain rule works. If $y = f(u)$ and $u = g(x)$Then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. This is a cancellation of the $du$ terms (formalised by limits).

Proof of the chain rule. Let $u = g(x)$ and define:

Then $\displaystyle\lim_{h \to 0} k(h) = 0$ and $f(u+h) - f(u) = [f'(u) + k(h)] \cdot h$ for all $h$.

Let $\Delta u = g(x+h) - g(x)$. Then:

Taking $h \to 0$: $\Delta u \to 0$ (since $g$ is differentiable, hence continuous), so $k(\Delta u) \to 0$Giving:

$\blacksquare$

Example

Find $\displaystyle\frac{d}{dx}\left[(3x^2 + 1)^5\right]$.

Let $u = 3x^2 + 1$So $y = u^5$.

Example

Find $\displaystyle\frac{d}{dx}\left[\sin^3(2x)\right]$.

Apply the chain rule twice:

Example

Find $\displaystyle\frac{d}{dx}\left[\ln(\sin x)\right]$.

Let $u = \sin x$So $y = \ln u$.

Example

Find $\displaystyle\frac{d}{dx}\left[e^{\sqrt{3x+1}}\right]$.

Apply the chain rule twice:

$\blacksquare$

Where we used $\displaystyle\lim_{h \to 0} \frac{e^h - 1}{h} = 1$.

Let $y = \ln x$So $e^y = x$. Differentiating implicitly: $e^y \frac{dy}{dx} = 1$So $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$. $\blacksquare$

Write $a^x = e^{x \ln a}$. Then:

$\blacksquare$

Derivation of $\arcsin x$. Let $y = \arcsin x$So $x = \sin y$. Differentiating: $1 = \cos y \cdot \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}}$.

Derivation of $\arctan x$. Let $y = \arctan x$So $x = \tan y$. Differentiating: $1 = \sec^2 y \cdot \frac{dy}{dx}$So $\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y$.

Since $\sec^2 y = 1 + \tan^2 y = 1 + x^2$We get $\cos^2 y = \frac{1}{1 + x^2}$So $\frac{dy}{dx} = \frac{1}{1 + x^2}$. $\blacksquare$

When $y$ is defined implicitly as a function of $x$Differentiate both sides with respect to $x$ And solve for $\frac{dy}{dx}$.

Example

Find $\displaystyle\frac{dy}{dx}$ for $x^2 + y^2 = 25$.

Differentiate both sides with respect to $x$:

Example

Find the equation of the tangent line to $x^2 + y^2 = 25$ at the point $(3, 4)$.

From the previous example, $\displaystyle\frac{dy}{dx} = -\frac{x}{y}$. At $(3, 4)$:

The tangent line is $y - 4 = -\frac{3}{4}(x - 3)$Or $3x + 4y = 25$.

Example

Find $\displaystyle\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$.

Differentiate implicitly:

Example

Find $\displaystyle\frac{d^2y}{dx^2}$ for $x^2 + y^2 = 25$.

We have $\frac{dy}{dx} = -\frac{x}{y}$.

Differentiating again:

Example

Find $\displaystyle\frac{dy}{dx}$ for $e^{xy} + y = x^2$.

Differentiate implicitly with respect to $x$:

The second derivative is the derivative of the first derivative:

Similarly, f'''(x)$$f^{(4)}(x)Etc. Denote higher-order derivatives.

• If f''(x) \gt 0$$f is concave up at $x$.
• If f''(x) \lt 0$$f is concave down at $x$.
• If $f''(x)$ changes sign at $x = c$Then $(c, f(c))$ is an inflection point.

Theorem. If $f$ has a local maximum at $c$ and $f''(c)$ exists, then $f''(c) \leq 0$. If $f$ has A local minimum at $c$ and $f''(c)$ exists, then $f''(c) \geq 0$.

This is the second derivative test for concavity at critical points.

Common notations for derivatives include:

• f'(x)$$f''(x)$$f'''(x) (prime notation)
• \frac{df}{dx}$$\frac{d^2f}{dx^2} (Leibniz notation)
• \dot{x}$$\ddot{x} (Newton’s dot notation for time derivatives)

When two or more quantities are related by an equation, their rates of change are also related.

Example

A ladder 10 ft long rests against a vertical wall. If the bottom slides away at 1 ft/s, how fast is The top sliding down when the bottom is 6 ft from the wall?

Let $x$ = distance from wall to bottom, $y$ = height of top on wall. Then $x^2 + y^2 = 100$.

Differentiate with respect to $t$:

When $x = 6$: $y = \sqrt{100 - 36} = 8$. Given $\frac{dx}{dt} = 1$:

The negative sign confirms the top is sliding down.

Example

A spherical balloon is inflated at a rate of 10 \mathrm{ cm^3/\mathrm{s. How fast is the radius Increasing when the radius is 5 cm? ($V = \frac{4}{3}\pi r^3$)

Example

A 15 ft ladder leans against a wall. The bottom slides away at 2 ft/s. How fast is the angle between The ladder and the ground changing when the bottom is 9 ft from the wall?

Let $\theta$ be the angle between the ladder and the ground. Then $\cos\theta = \frac{x}{15}$So $x = 15\cos\theta$.

Differentiating with respect to $t$:

When $x = 9$: $\cos\theta = \frac{9}{15} = \frac{3}{5}$So $\sin\theta = \frac{4}{5}$.

The angle is decreasing at $\frac{1}{6}$ rad/s.

The tangent line at $x = a$ gives a linear approximation:

This is useful for estimating values of functions near known points.

Example

Use linear approximation to estimate $\sqrt{4.1}$.

Let f(x) = \sqrt{x}$$a = 4.

(The actual value is approximately 2.02485.)

Example

Use linear approximation to estimate $(1.02)^{10}$.

Let f(x) = x^{10}$$a = 1.

(The actual value is approximately 1.21899, showing that linear approximation is less accurate when $x$ is farther from $a$.)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$Then there exists at least one $c \in (a, b)$ such that:

This means the instantaneous rate of change equals the average rate of change at some interior Point.

Geometric interpretation. The MVT guarantees the existence of a tangent line parallel to the Secant line joining $(a, f(a))$ and $(b, f(b))$.

Proof of the MVT. Let $g(x) = f(x) - \frac{f(b) - f(a)}{b-a}(x - a) - f(a)$. Then $g(a) = g(b) = 0$. By Rolle’s Theorem, there exists $c \in (a, b)$ with $g'(c) = 0$Which gives $f'(c) = \frac{f(b) - f(a)}{b-a}$. $\blacksquare$

If $f$ is continuous on $[a, b]$Differentiable on $(a, b)$And $f(a) = f(b)$Then there exists At least one $c \in (a, b)$ such that $f'(c) = 0$.

Rolle’s Theorem is a special case of the MVT where $f(a) = f(b)$.

Application of Rolle’s Theorem. Rolle’s Theorem is used to prove that a differentiable function Has at most one root in an interval. If it had two roots, Rolle’s Theorem would give a point where The derivative is zero, leading to a contradiction.

Example

Show that $f(x) = x^3 - 3x + 1$ has exactly one real root.

Note that $f(0) = 1$ and $f(-2) = -8 + 6 + 1 = -1$. By the IVT, there is at least one root in $(-2, 0)$.

Suppose there are two distinct roots $a \lt b$. Then $f(a) = f(b) = 0$And by Rolle’s Theorem, There exists $c \in (a, b)$ with $f'(c) = 0$. But $f'(x) = 3x^2 - 3 = 3(x+1)(x-1)$Which is zero Only at $x = \pm 1$. This does not lead to a contradiction on its own, so we must examine further.

In fact, f(-2) = -1$$f(0) = 1$$f(1) = -1$$f(2) = 3. So $f$ has at least one root in $(-2, 0)$At least one in $(0, 1)$And possibly one in $(1, 2)$. Rolle’s Theorem applied to Consecutive roots shows that each pair of roots must bracket either $x = -1$ or $x = 1$.

Testing: $f(-1) = -1 + 3 + 1 = 3 \gt 0$. So the root in $(-2, 0)$ is in $(-2, -1)$And the root in $(0, 1)$ is in $(0, 1)$. These bracket different critical points. In fact, $f$ has three real roots, And Rolle’s Theorem confirms this is consistent since the critical points at $x = \pm 1$ bracket the Three roots.

A critical number of $f$ is a value $c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ does Not exist.

Let $c$ be a critical number:

• If $f'$ changes from positive to negative at $c$Then $f(c)$ is a local maximum.
• If $f'$ changes from negative to positive at $c$Then $f(c)$ is a local minimum.
• If $f'$ does not change sign at $c$Then $f(c)$ is neither.

Let $f''(c)$ exist with $f'(c) = 0$:

• If $f''(c) \gt 0$Then $f(c)$ is a local minimum.
• If $f''(c) \lt 0$Then $f(c)$ is a local maximum.
• If $f''(c) = 0$The test is inconclusive.

When to use which test. The first derivative test always works. The second derivative test is Faster but sometimes inconclusive. When $f''(c) = 0$You must fall back to the first derivative Test.

Example where the second derivative test is inconclusive. Consider $f(x) = x^4$. $f'(x) = 4x^3 = 0$ gives $x = 0$And $f''(x) = 12x^2$So $f''(0) = 0$ — the test is inconclusive. But $f'(x) = 4x^3$ changes from negative to positive at $x = 0$So $f$ has a local minimum at $x = 0$ by the first derivative test.

To solve optimization problems:

1. Identify the quantity to optimize and the constraint.
2. Write an equation relating the variables.
3. Express the quantity as a function of a single variable.
4. Find critical points and apply the first or second derivative test.
5. Verify the result answers the original question.

Example

A farmer has 200 ft of fencing. Find the dimensions of the rectangular pen that maximizes the area.

Let $x$ and $y$ be the dimensions. Then $2x + 2y = 200$So $y = 100 - x$.

So x = 50$$y = 50And the maximum area is 2500 \mathrm{ ft^2.

Example

Find the minimum value of $f(x) = x + \frac{4}{x}$ for $x \gt 0$.

The minimum value is $4$ at $x = 2$.

Example

A rectangular box with a square base has volume 500 \mathrm{ cm^3. The material for the base costs Twice as much as the material for the sides. Minimise the cost of the material.

Let the base have side length $x$ cm and height $h$ cm. Let the cost per unit area of the sides be $c$. Then the base costs $2c$ per unit area.

Volume: $x^2 h = 500$So $h = \frac{500}{x^2}$.

Cost: $C = 2c \cdot x^2 + 4c \cdot xh = 2cx^2 + 4cx \cdot \frac{500}{x^2} = 2cx^2 + \frac{2000c}{x}$.

The cost is minimised when the box is a cube. $\frac{d^2C}{dx^2} = 4c + \frac{4000c}{x^3} \gt 0$ Confirming a minimum.

When the domain is a closed interval $[a, b]$The absolute maximum and minimum occur at critical Points or at endpoints.

Procedure:

1. Find all critical points in $(a, b)$.
2. Evaluate $f$ at all critical points and at both endpoints.
3. The largest value is the absolute max; the smallest is the absolute min.

Example

Find the absolute maximum and minimum of $f(x) = x^3 - 3x^2 + 4$ on $[-1, 4]$.

f(-1) = -1 - 3 + 4 = 0$$f(0) = 4$$f(2) = 8 - 12 + 4 = 0$$f(4) = 64 - 48 + 4 = 20.

Absolute minimum: $0$ at $x = -1$ and $x = 2$. Absolute maximum: $20$ at $x = 4$.