Chapter 3: Composite, Implicit, and Inverse Function Differentiation

Find $\dfrac{d}{dx}\left[\cos^3(x)\right]$.

Solution. Rewrite as $[\cos(x)]^3$. Here the outer function is $u^3$ and the inner function is $\cos x$.

Apply the chain rule: $$\frac{d}{dx}[\cos(x)]^3 = 3[\cos(x)]^2 \cdot (-\sin x) = -3\cos^2(x)\sin(x).$$

The key insight is recognizing that the exponent $3$ is the outermost operation acting on $\cos x$. The power rule gives $3[\cos x]^2$, and then we multiply by the derivative of the inside, which is $-\sin x$.

Find $\dfrac{d}{dx}\left[e^{3x^2 + 1}\right]$.

Solution. The outer function is $e^u$ and the inner function is $u = 3x^2 + 1$.

Since $\frac{d}{du}e^u = e^u$, we have $$\frac{d}{dx}\left[e^{3x^2+1}\right] = e^{3x^2+1} \cdot \frac{d}{dx}(3x^2+1) = e^{3x^2+1} \cdot 6x = 6x\,e^{3x^2+1}.$$

The exponential function has the special property that its derivative is itself, so the outer derivative simply reproduces the original exponential. All the action comes from the inner derivative $6x$.

Find $\dfrac{d}{dx}\left[\ln(\sin(e^x))\right]$.

Solution. There are three nested functions here. Identify them from outside to inside:

• Outermost: $\ln(u)$ with derivative $\frac{1}{u}$.
• Middle: $\sin(v)$ with derivative $\cos(v)$.
• Innermost: $e^x$ with derivative $e^x$.

Apply the chain rule layer by layer: $$\frac{d}{dx}\left[\ln(\sin(e^x))\right] = \frac{1}{\sin(e^x)} \cdot \cos(e^x) \cdot e^x.$$

Simplify using $\frac{\cos\theta}{\sin\theta} = \cot\theta$: $$= e^x \cot(e^x).$$

Each function in the chain contributes one factor to the derivative. Practice identifying the layers, and the computation becomes mechanical.

Find $\dfrac{d}{dx}\left[\ln(x^3 + 5)\right]$.

Solution. The outer function is $\ln(u)$ with derivative $\frac{1}{u}$, and the inner function is $u = x^3 + 5$.

Apply the chain rule: $$\frac{d}{dx}\left[\ln(x^3+5)\right] = \frac{1}{x^3+5} \cdot 3x^2 = \frac{3x^2}{x^3+5}.$$

A useful pattern emerges: $\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$ for any differentiable $g > 0$. This rule appears constantly in AP Calculus, particularly in related-rates and implicit differentiation problems involving logarithms.

Find $\dfrac{d}{dx}\left[x^2 \cos(x^3)\right]$.

Solution. This is a product of $x^2$ and $\cos(x^3)$. Apply the product rule first: $$\frac{d}{dx}\left[x^2 \cos(x^3)\right] = 2x\cos(x^3) + x^2 \cdot \frac{d}{dx}[\cos(x^3)].$$

Now handle $\frac{d}{dx}[\cos(x^3)]$ with the chain rule: outer $\cos(u)$, inner $u = x^3$: $$\frac{d}{dx}[\cos(x^3)] = -\sin(x^3) \cdot 3x^2 = -3x^2\sin(x^3).$$

Combine: $$\frac{d}{dx}\left[x^2\cos(x^3)\right] = 2x\cos(x^3) + x^2(-3x^2\sin(x^3)) = 2x\cos(x^3) - 3x^4\sin(x^3).$$

The rule of thumb: when the chain rule and another rule (product, quotient) are both needed, apply the outer rule first, then handle any composite pieces with the chain rule.

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

Solution. Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$ $$2x + 2y\,\frac{dy}{dx} = 0.$$

Solve for $\frac{dy}{dx}$:

$$2y\,\frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = -\frac{x}{y}.$$

This result makes geometric sense. At the point $(3, 4)$ on the circle, the slope of the tangent line is $-\frac{3}{4}$, and the radius to that point has slope $\frac{4}{3}$. The tangent is perpendicular to the radius, exactly as Euclidean geometry predicts for a circle.

Find the equation of the tangent line to $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ at the point $(x_0, y_0) = \left(\frac{3\sqrt{2}}{2}, \sqrt{2}\right)$.

Solution. First find $\frac{dy}{dx}$. Differentiate implicitly:

$$\frac{2x}{9} + \frac{2y}{4} \cdot \frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = -\frac{2x/9}{2y/4} = -\frac{2x}{9} \cdot \frac{4}{2y} = -\frac{4x}{9y}.$$

Evaluate at the given point:

$$\frac{dy}{dx}\bigg|_{\left(\frac{3\sqrt{2}}{2},\,\sqrt{2}\right)} = -\frac{4 \cdot \frac{3\sqrt{2}}{2}}{9\sqrt{2}} = -\frac{6\sqrt{2}}{9\sqrt{2}} = -\frac{2}{3}.$$

The tangent line is:

$$y - \sqrt{2} = -\frac{2}{3}\left(x - \frac{3\sqrt{2}}{2}\right).$$

Simplifying: $y = -\frac{2}{3}x + \sqrt{2} + \sqrt{2} = -\frac{2}{3}x + 2\sqrt{2}.$

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

Solution. This famous curve, the folium of Descartes, cannot be solved explicitly for $y$ in closed form. Implicit differentiation is the only practical approach.

Differentiate both sides with respect to $x$:

$$3x^2 + 3y^2\,\frac{dy}{dx} = 6y + 6x\,\frac{dy}{dx}.$$

On the right side, we used the product rule on $6xy$: the derivative of $6xy$ with respect to $x$ is $6\left(y + x\frac{dy}{dx}\right) = 6y + 6x\frac{dy}{dx}$.

Collect terms with $\frac{dy}{dx}$:

$$3y^2\,\frac{dy}{dx} - 6x\,\frac{dy}{dx} = 6y - 3x^2$$ $$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$ $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}.$$

Notice that the derivative is undefined when $y^2 = 2x$, which corresponds to points where the tangent line is vertical. At the origin $(0,0)$, the formula gives $\frac{0}{0}$, which is consistent with the fact that the curve crosses itself there and has two distinct tangent lines.

Find $\dfrac{dy}{dx}$ for $x\sin(y) = \cos(x + y)$.

Solution. Differentiate both sides with respect to $x$. The left side requires the product rule; the right side requires the chain rule.

Left side: $\dfrac{d}{dx}[x\sin y] = \sin y + x\cos(y)\dfrac{dy}{dx}$.

Right side: $\dfrac{d}{dx}[\cos(x+y)] = -\sin(x+y)\cdot\left(1 + \dfrac{dy}{dx}\right)$.

Setting them equal and collecting $\dfrac{dy}{dx}$ terms:

$$\sin y + x\cos(y)\frac{dy}{dx} = -\sin(x+y) - \sin(x+y)\frac{dy}{dx}$$ $$\frac{dy}{dx}\bigl[x\cos y + \sin(x+y)\bigr] = -\sin(x+y) - \sin y$$ $$\frac{dy}{dx} = \frac{-\sin(x+y) - \sin y}{x\cos y + \sin(x+y)}.$$

Find all points on the curve $x^2 - xy + y^2 = 3$ where the tangent line is horizontal.

Solution. Differentiate implicitly: $$2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$$ $$\frac{dy}{dx}(2y - x) = y - 2x$$ $$\frac{dy}{dx} = \frac{y - 2x}{2y - x}.$$

For a horizontal tangent, set $\dfrac{dy}{dx} = 0$: the numerator must be zero (with non-zero denominator), so $y - 2x = 0$, giving $y = 2x$.

Substitute $y = 2x$ into the original equation:

$$x^2 - x(2x) + (2x)^2 = 3 \implies x^2 - 2x^2 + 4x^2 = 3 \implies 3x^2 = 3 \implies x = \pm 1.$$

The two points of horizontal tangency are $\boldsymbol{(1,\,2)}$ and $\boldsymbol{(-1,\,-2)}$. (Verify: at $(1,2)$, the denominator $2y-x = 4-1 = 3 \neq 0$. ✓)

Let $f$ be a differentiable function with $f(2) = 5$ and $f'(2) = 3$. Find $(f^{-1})'(5)$.

Solution. By Theorem 3.2:

$$(f^{-1})'(5) = \frac{1}{f'(f^{-1}(5))}.$$

Since $f(2) = 5$, we know $f^{-1}(5) = 2$. Therefore:

$$(f^{-1})'(5) = \frac{1}{f'(2)} = \frac{1}{3}.$$

This type of problem appears regularly on AP Calculus exams, often presented as a table of values. The key is to first determine $f^{-1}(a)$ by reading the table in reverse, then evaluate $f'$ at that point and take the reciprocal.

Find $\dfrac{d}{dx}\left[\arctan(3x^2)\right]$.

Solution. Combine the inverse tangent derivative with the chain rule:

$$\frac{d}{dx}\left[\arctan(3x^2)\right] = \frac{1}{1+(3x^2)^2} \cdot \frac{d}{dx}(3x^2) = \frac{6x}{1 + 9x^4}.$$

Whenever an inverse trigonometric function has a non-trivial argument, the chain rule is required. The derivative of the outer function ($\arctan$) is $\frac{1}{1+u^2}$, and we multiply by the derivative of the inner function ($6x$).

Find $\dfrac{d}{dx}\left[\arccos(\sqrt{x})\right]$.

Solution. Apply the chain rule. The outer function is $\arccos(u)$ with derivative $-\dfrac{1}{\sqrt{1-u^2}}$, and the inner function is $u = \sqrt{x} = x^{1/2}$ with derivative $\dfrac{1}{2\sqrt{x}}$.

$$\frac{d}{dx}\left[\arccos(\sqrt{x})\right] = -\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{2\sqrt{x(1-x)}}.$$

This is valid for $0 < x < 1$. The negative sign reflects the fact that $\arccos$ is a decreasing function.

Find $\dfrac{d}{dx}\left[x\arctan(x)\right]$.

Solution. Apply the product rule: $$\frac{d}{dx}[x\arctan(x)] = \arctan(x) + x \cdot \frac{1}{1+x^2} = \arctan(x) + \frac{x}{1+x^2}.$$

This combination — $\arctan(x) + \dfrac{x}{1+x^2}$ — appears when integrating $\arctan(x)$ by parts and is worth recognizing. Note that as $x \to \infty$, the term $\frac{x}{1+x^2} \to 0$, so the derivative approaches $\frac{\pi}{2}$, consistent with the fact that $x\arctan(x) \approx \frac{\pi}{2}x$ for large $x$.

Compute $\dfrac{d}{dx}\left[\arcsin(x) + \arccos(x)\right]$ and explain the result.

Solution. $$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}.$$

The sum of derivatives is $\dfrac{1}{\sqrt{1-x^2}} - \dfrac{1}{\sqrt{1-x^2}} = 0$.

A derivative of zero means the function is constant. Indeed, the identity $\arcsin(x) + \arccos(x) = \dfrac{\pi}{2}$ holds for all $x \in [-1, 1]$: since $\sin\theta = \cos(\frac{\pi}{2} - \theta)$, if $\sin\theta = x$ then $\arccos(x) = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \arcsin(x)$. This is a beautiful illustration of why the derivative of a constant is zero.

Given $x^2 + y^2 = 25$, find $\dfrac{d^2y}{dx^2}$.

Solution. From Example 3.6, we found $\frac{dy}{dx} = -\frac{x}{y}$. Differentiate again using the quotient rule:

$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2}.$$

Substitute $\frac{dy}{dx} = -\frac{x}{y}$:

$$= -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{\frac{y^2 + x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3}.$$

Since $x^2 + y^2 = 25$, this simplifies to:

$$\frac{d^2y}{dx^2} = -\frac{25}{y^3}.$$

At the point $(3, 4)$: $\frac{d^2y}{dx^2} = -\frac{25}{64}$. The negative sign confirms that the upper semicircle is concave down, which matches our geometric intuition.

Let $f(x) = x^5 - 4x^3 + 2x$. Find $f^{(4)}(x)$ (the fourth derivative).

Solution. Differentiate successively:

• $f'(x) = 5x^4 - 12x^2 + 2$
• $f''(x) = 20x^3 - 24x$
• $f'''(x) = 60x^2 - 24$
• $f^{(4)}(x) = 120x$

Observe that each differentiation reduces the degree by one. A degree-$n$ polynomial has its $(n+1)$-th derivative equal to zero. Here the fifth derivative would be $120$, and the sixth derivative would be $0$.

Find $y''$ for $y = \sin(3x)$ and describe the relationship between $y''$ and $y$.

Solution. Apply the chain rule twice:

$$y' = 3\cos(3x), \qquad y'' = -9\sin(3x).$$

Notice that $y'' = -9\sin(3x) = -9y$. The function satisfies the differential equation $y'' + 9y = 0$, which governs simple harmonic oscillation. Any function of the form $y = A\sin(3x) + B\cos(3x)$ satisfies this equation, explaining the fundamental role of sinusoidal functions in physics.

Let $y = e^{ax}$ for a constant $a$. Find a formula for $y^{(n)}$ (the $n$-th derivative).

Solution. Compute the first several derivatives:

$$y' = ae^{ax}, \quad y'' = a^2e^{ax}, \quad y''' = a^3e^{ax}, \quad \ldots$$

The pattern is clear: $y^{(n)} = a^n e^{ax}$.

Proof by induction: Base case $n=1$: $y' = ae^{ax}$ ✓. Inductive step: if $y^{(k)} = a^k e^{ax}$, then $y^{(k+1)} = \frac{d}{dx}[a^k e^{ax}] = a^k \cdot a e^{ax} = a^{k+1}e^{ax}$. ✓

In particular, $e^x$ has the remarkable property that every derivative equals $e^x$ (i.e., $a = 1$ gives $y^{(n)} = e^x$ for all $n$).

A particle's position (in meters) is $s(t) = t^3 - 6t^2 + 9t$ for $t \ge 0$ (seconds).

(a) Find velocity $v(t)$ and acceleration $a(t)$.
(b) When is the particle momentarily at rest?
(c) When is the speed increasing?

Solution.

(a) $$v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3), \qquad a(t) = v'(t) = 6t - 12 = 6(t-2).$$

(b) $v(t)=0$ at $t = 1$ and $t = 3$.

(c) Speed increases when velocity and acceleration have the same sign:

• $t \in (0,1)$: $v > 0$, $a < 0$ — speed decreasing.
• $t \in (1,2)$: $v < 0$, $a < 0$ — speed increasing (moving left faster).
• $t \in (2,3)$: $v < 0$, $a > 0$ — speed decreasing.
• $t \in (3,\infty)$: $v > 0$, $a > 0$ — speed increasing.

Speed is increasing for $t \in (1,2) \cup (3, \infty)$.