4.5 The Chain Rule and Implicit Differentiation

4.5.1 Preliminary Knowledge: Composition of Functions

Definition 1
Given functions \(f\) and \(g\), the composition \(f\circ g\) of \(f\) and \(g\) is defined by
\[D_{f \circ g} = \{ x \in {D_g}:\;\,g(x) \in {D_f}\} \;\;\text{ and } \;\; (f \circ g)(x) = f(g(x)), \;\; \forall \; x \in {D_{f \circ g}}.\]

Note that:
\[x\in D_{f \circ g} \Longleftrightarrow x \in D_g \;\; \text{and} \;\; g(x) \in D_f.\]

Example 1
Express each function \(h\) as the composition of two nontrivial functions \(f\) and \(g\), such that \(h = f\circ g\) and find \(D_{f\circ g}\).

1. \(\displaystyle h(x)= \frac{1}{\sqrt{x^2-9}}\);
2. \(h(x)=\arcsin(\ln x^3)\);
3. \(h(x)=\tan(\arccos x)\);
4. \(h(x) = \ln^2\sqrt[4]{x^3}\).

Solution

(a) Let \(g(x)=x^2-9\) and \(f(x)=x^{-1/2}\), then
\[(f\circ g)(x)= f(g(x))=f(x^2-9)=(x^2-9)^{-1/2}=\frac{1}{\sqrt{x^2-9}}= h(x).\]
Since \(D_g=\mathbb{R}\), it follows that
\[x\in D_{f\circ g} \Longleftrightarrow x^2-9 > 0\Longleftrightarrow x \in (-\infty, -3)\cup(3, \infty).\]

(b) First, note that
\[h(x)=\arcsin(\ln x^3) = \arcsin(3\ln x) \Longrightarrow g(x) = 3\ln x \text{ and } f(x) =\arcsin x\]
(verify that \(h = f\circ g\)). As for the domain, we have
\[x\in D_{f\circ g} \underset{?}{\Longleftrightarrow} x > 0 \text{ and } -1 \leq 3\ln x \leq 1 \underset{?}{\Longleftrightarrow} x\in \left [\frac{1}{\sqrt[3]{e}}, \sqrt[3]{e}\right ].\]

(c) Let \(g(x)=\arccos x\) and \(f(x)=\tan x\), then \(h = f\circ g\).
\[x\in D_{f\circ g} \underset{?}{\Longleftrightarrow} -1\leq x \leq 1 \text{ and } -\frac{\pi}{2} < \arccos x < \frac{\pi}{2}.\]
Since, by definition, \(0 \leq \arccos x \leq \pi\), and \(\arccos x\) is decreasing
\[x\in D_{f\circ g} \underset{?}{\Longleftrightarrow} x\in \left [0,1\right ].\]

(d) First note that
\[\ln^2 \sqrt[4]{x^3} = \left (\ln x^{3/4}\right )^2 = \left (\frac{3}{4}\ln x \right )^2 \Longrightarrow g(x) = \frac{3}{4}\ln x \text{ and } f(x) = x^2,\]
that is,
\[(f\circ g)(x)= f(g(x))=f\left (\frac{3}{4}\ln x\right )=\left(\frac{3}{4}\ln x\right )^2=h(x).
\Longrightarrow x\in D_{f\circ g} \underset{?}{\Longleftrightarrow} x \in (0, \infty).\]

Example 2

Express each function \(F\) as the composition of two nontrivial functions \(f\), \(H\), in such a way that \(F = f\circ H\) and \(H = g \circ h\), with \(g\) and \(h\) nontrivial.

(a) \(F(x)=\arctan^3 \left(\cot 2x\right)\);

(b) \(F(x)=e^{\sin \pi x^2}\);

(c) \(F(x)=\sqrt[3]{\tan(3x^2+5)}\);

(d) \(\displaystyle F(x)= \ln\left (\frac{1}{1+ x^2 }\right )^3\).

Solution

(a)
Let \(f(x)=x^3\) and \(H(x)=\arctan \left(\cot 2x\right)\), then
\[(f\circ H)(x) = f((H(x)) = f\left (\arctan \left(\cot 2x\right)\right ) = \left (\arctan \left(\cot 2x\right)\right )^3=F(x),\]

and, if \(g(x)= \arctan x\) and \(h(x)= \cot 2x\), then
\[(g\circ h)(x) = g(h(x))= \arctan (\cot 2x) = H(x),\]

thus
\[F(x) = (f\circ H)(x) = (f\circ (g\circ h))(x).\]

(b) \(f(x)=e^x\), \(g(x)= \sin x\), \(h(x)= \pi x^2 \);

(c) \(f(x)=\sqrt[3]{x}\); \(g(x)= \tan x \); \(h(x)= 3x^2+5\);

(d) \(\displaystyle f(x)= \ln x\), \(g(x)=^{-3}\), \(h(x) = 1+ x^2\). However, note that
\[F(x)= \ln\left (\frac{1}{1+ x^2 }\right )^3 = 3 \ln \frac{1}{1+ x^2 } =-3\ln(1+x^2),\]
and we only need
\[f(x)=-3\ln x, \;\; g(x) = 1+x^2, \text{ and } F(x) = (f\circ g)(x).\]

4.5.3 The Chain Rule

We begin by interpreting the examples of the previous section in the context of composition of functions. This, in turn, will lead to the statement of the chain rule
\[
\renewcommand{\arraystretch}{2.3}
\begin{array}{c|c}
h(x) &h'(x) \\ \hline
\csc x^5 & -5x^4\cot x^5 \csc x^5 \\ \hline
\sec \sqrt[3]{x} &\displaystyle \frac{1}{3}\frac{1}{\sqrt[3]{x^2}}\sec\sqrt[3]{x}\tan\sqrt[3]{x} \\\hline
\cos^4 x & – 4\cos^3 x\sin x \\ \hline
\sqrt{\csc x^5} & \displaystyle -5x^4\cot x^5 \csc x^5\cdot\frac{1}{2\sqrt{\csc x^5}}
\end{array}
\]

The following table contains an analysis of these examples, thinking of each \(h\) as a composition of functions \(h(x)=f(g(x))\). We will use the notation \(u=g(x)\).
\[
\renewcommand{\arraystretch}{2.3}
\begin{array}{c||c|c||c|c||c}
f(g(x)) &f(u) &f'(u) &u=g(x) &u’=g'(x) &f'(u)g'(x) \\ \hline
\csc x^5 & \csc u & -\csc u \cot u & x^5 & 5x^4 & ( -\csc u \cot u )(5x^4) \\ \hline
\sec \sqrt[3]{x} &\sec u &\sec u \tan u & \sqrt[3]{x} &\displaystyle \frac{1}{3}\frac{1}{\sqrt[3]{x^2}} &\displaystyle (\sec u \tan u )\left(\frac{1}{3}\frac{1}{\sqrt[3]{x^2}} \right) \\\hline
\cos^4 x & u^4 & 4u^3 & \cos x & -\sin x & 4u^3(-\sin x)\\ \hline
\sqrt{\csc x^5} &\sqrt u &\displaystyle \frac{1}{2}\frac{1}{\sqrt u} & \csc x^5 & \displaystyle -5x^4\cot x^5 \csc x^5 & \displaystyle \frac{1}{2}\frac{1}{\sqrt u}( -5x^4\cot x^5 \csc x^5 )
\end{array}
\]
We note that if we replace \(u=g(x)\) in the last column, then we obtain exactly the same expression for the derivative as obtained from the definition. Namely \(h'(x)=f'(g(x))g'(x)\). It turns out that this true in general and is the content of the chain rule

Theorem 1. (The Chain Rule) (4.4) If \(f\) and \(g\) are both differentiable functions then

\[ (f\circ g)'(x)= f'(g(x)) g'(x) \text{ for all } x \in D_{f\circ g}\tag{4.4}\]

For a proof of this rule see. In this and the next two sections we are going to use the Chain Rule in different contexts.

A way to visualize the Chain Rule is to draw the chain of functions, and the derivatives, at each stage of the chain, below the arrows, as in the following diagram.

\[\begin{array}{ccccc}
x & \mapsto & u=g(x) & \mapsto & y=f(u)=f(g(x)) \\
& \frac{d u}{d x}=g'(x) & & \frac{d y}{d u}=f'(u)=f'(g(x)) &
\end{array}\]

The chain rule then can be written, using the different notations, as the product of the rates of change at each stage:
\[(f\circ g)'(x)= f'(g(x)) g'(x) \text{ or } \frac{d y}{d x}=\frac{d y}{d u}\frac{d u}{d x}.\]

Example 1. Let \(a \in \mathbb{R}\) be nonzero. Use the Chain Rule to compute the derivative of (a) \( \sin a x\); (b) \( \tan x^2\); (c) \(\sqrt[3]{\csc^2 a x}\), \(a \neq 0\) constant.

Solution
(a) Let \(u=g(x)=a x\) and \(f(u)=\sin u \Longrightarrow\)
\[\begin{array}{ccccc}
x & \mapsto & u=ax & \mapsto & y=\sin u =e\sin ax \\
& g'(x)=a & & f'(u)=\cos u & \\
\end{array}\]

\[\Longrightarrow (f\circ g)'(x) = g'(x)f'(u) = a \, \cos u \Longrightarrow (f\circ g)'(x)= g'(x)f'(g(x)) = a \cos ax.\]

(b) Let \(u=g(x)= x^2\) and \(f(u)=\tan u \Longrightarrow \)
\[\begin{array}{ccccc}
x & \mapsto & u=x^2 & \mapsto & y=\tan u =\tan x^2 \\
& g'(x)=2x & & f'(u)=-\csc^2 u & \\
\end{array}\]

\[\Longrightarrow (f\circ g)'(x) = g'(x)f'(u) = -2x \csc^2 u \Longrightarrow (f\circ g)'(x)= g'(x)f'(g(x)) = -2x \csc x^2.\]

(c) Since \(\sqrt[3]{\csc^2 a x} = \csc^{2/3} a x \), letting \(u=g(x)= \csc ax\) and \(f(u)=\ u^{2/3} \Longrightarrow \)
\[\begin{array}{ccccc}
x & \mapsto & u=\csc ax & \mapsto & f(u) =u^{2/3} = (\csc x)^{2/3} = \sqrt[3]{\csc^2x} \\
& g'(x)=- \csc ax\cot ax & & f'(u)=\frac{2}{3} u^{-1/3} & \\
\end{array}\]

\[\Longrightarrow (f\circ g)'(x) = g'(x)f'(u) = – \csc ax\cot ax \cdot \frac{2}{3} u^{-1/3} \Longrightarrow\]
\[ \Longrightarrow (f\circ g)'(x)= g'(x)f'(g(x)) = – \csc ax\cot ax \cos ax.\]

Example 2. Find the \(x\)-coordinate(s) of the point(s) on the graph of \(\displaystyle F(x)=\frac{x^2}{\sqrt{x-2\sqrt{x}}}\) whose tangent line is horizontal.

Solution
First, we apply the quotient rule:
\[F'(x) = \frac{
\sqrt{x-2\sqrt{x}}
\left (2x\right ) -x ^2\left (\sqrt{x-2\sqrt{x}}\right )’}
{x-2\sqrt{x} }.\]
Next, we apply the Chain Rule to compute the derivative of \(h(x) = \sqrt{x-2\sqrt{x}}\):
\[u=g(x)=x-2\sqrt{x}, \;\; f(u)=\sqrt u \Longrightarrow \]

\[\begin{array}{ccccc}
x & \mapsto & u=x-2\sqrt{x} & \mapsto & f(u) = \sqrt u=\sqrt{x-2\sqrt{x}} \\
& g'(x)=1 – \frac{1}{\sqrt{x}}& & f'(u)=\frac{1}{2\sqrt u} & \\
\end{array}
\]
hence
\[ (f\circ g)'(x) = g'(x)f'(u) = \left (1 – \frac{1}{\sqrt{x}}\right )\cdot \frac{1}{2\sqrt u}
= \left (1 – \frac{1}{\sqrt{x}}\right )\cdot \frac{1}{2\sqrt {x-2\sqrt{x}}}
=\frac{ \sqrt{x}-1}{2\,\sqrt{x}\,\sqrt{x-2\sqrt{x}}}.
\]
Thus
\begin{align*}
F'(x) &= \frac{\displaystyle
2x \,\sqrt{x-2\sqrt{x}}
– x^2
\frac{ \sqrt{x}-1}{2\,\sqrt{x}\,\sqrt{x-2\sqrt{x}}}}
{x-2\sqrt{x} } = \frac{4\sqrt{x}\, x \left (x-2\sqrt{x}\right )- x^2\left (\sqrt{x}-1\right )}{2\,\sqrt{x}\,\sqrt{x-2\sqrt{x}}\left (x-2\sqrt{x}\right )}\\
& \\
&= \frac{4\sqrt{x}\, x^2 -8x^2- x^2\,\sqrt{x}+x^2}{2\,\sqrt{x-2\sqrt{x}}\left (x\,\sqrt{x}-2x\right )}
\underset{?}{=} \frac{ x \left (3\sqrt{x}-7\right )}{2\,\sqrt{x-2\sqrt{x}}\left ( \sqrt{x}-2 \right )}.
\end{align*}

For a horizontal tangent line
\[F'(x) = 0 \;\;\Longleftrightarrow \;\; x \left (3\sqrt{x}-7\right ) = 0 \;\; \text{ and } \;\; 2\,\sqrt{x-2\sqrt{x}}\left ( \sqrt{x}-2 \right ) \neq 0 \;\; \underset{?}{\Longleftrightarrow} \;\; x= \frac{49}{9}.\]

4.5.5 Extensions of the Chain Rule

We begin by computing the derivative of a function of the form
\[F(x) = (f\circ g\circ h)(x) = f(g(h(x))).\]
To accomplish this, it is necessary to apply the Chain Rule twice.

Let \(H(x) = (g\circ h )(x)= g(h(x)) \), then
\[F(x) = (f\circ H)(x) \Longrightarrow F'(x) = f'(H(x))H'(x)= f'(g(h(x))H'(x).\]

Applying the Chain Rule to compute \(H’\),
\[F'(x) = f'(g(h(x))g'(h(x))h'(x).\]

Remark: Rather than memorizing this formula, what is more significant is to understand the process, as it can be applied to functions that involve more than three compositions, as the example below shows.

Compute the derivative of (a) \(F(x) = \sin^4(x^3+1)\); (b) \(G(x) = x^3\tan^5\sqrt[4]{x^2+x}\)

Solution
(a) Note that \(F(x) = \sin^4(x^3+1) = \left (\sin (x^3+1)\right )^4\). Set \(H(x) =\sin (x^3+1)\), and
\[F'(x) = 4(H(x))^3H'(x) = 4 \sin^3(x^3+1)\cdot \cos (x^3+1) \cdot 3x^2.\]

(b) Since \(G(x) = x^3\tan^5\sqrt[4]{x^2+x} = x^3\left (\tan \left (x^2+x\right )^{1/4}\right )^5\), applying the Product Rule and the Chain Rule
\begin{align*}
G'(x) &= 3x^2\tan^5\sqrt[4]{x^2+x}\, +\\
& \ + x^3\cdot
5\left (\tan \left (x^2+x\right )^{1/4}\right )^4\cdot \sec^2\left (x^2+x\right )^{1/4}\cdot\frac{1}{4}\left (x^2+x\right )^{-3/4}\cdot(2x+1)=\\
&=
\frac{1}{4}x^2\tan^4\sqrt[4]{x^2+x} \left (12 \tan\sqrt[4]{x^2+x} +\frac{5x(2x+1)}{\sqrt[4]{(x^2+x)^3}}\sec^2\sqrt[4]{x^2+x}\right ) .
\end{align*}

4.5.7 Implicit Differentiation

An immediate application of the chain rule consist in finding slopes of tangent line to curves in the \(xy\)-plane defined by implicit equations. This method is called Implicit Differentation

Example 1. (Implicit Differentiation Method) The graph of the equation \[x^2 + 6xy + y^2=-4\] is a hyperbola (see Figure 4.2). (a) Verify that \(P\left (-1, 1 \right )\) is on the graph; (b) find the slope-intercept equation of the line tangent to the graph at \(P\).

Figure 4.2 Example (a)

Figure 4.2 Example (b)

(a) Substituting the coordinates of \(P\) into the equation: \(1^2 + 6(1)(-1) +(-1)^2=-4\). That is, the coordinates satisfy the equation. Thus, \(P\) is a point on the hyperbola.

(b) To find the slope of the tangent line to the graph, we can proceed in two ways:

(i) solve for \(y\) to obtain (via the quadratic formula)
\[f(x) = -3x – 2\sqrt{2x^2-1} \; \text{or} \; g(x) = -3x+ 2\sqrt{2x^2-1}.\]
Figure 4.2(b) shows the hyperbola divided in two parts of different colors. Each part of each color corresponds to either the graph of \(f\) or the graph of \(g\). Since we are looking for the function whose graph passes through \(P(-1,1)\), this function is given by
\[f(x) = -3x – 2\sqrt{2x^2-1}, \;\; x\in \left (-\infty, -\frac{1}{\sqrt 2}\right ) \cup \left (\frac{1}{\sqrt 2}, \infty\right ).\]
A direct application of the chain rule yields
\[f'(x) = -3-\frac{4x}{\sqrt{2x^2-1}}\;\; \Longrightarrow f'(1) = 1.\]
It follows that the tangent line to the hyperbola at \(P\) has equation
\[y-1=1\cdot(x+1) \Longleftrightarrow y = x +2.\]

(ii) Implicit Differentiaton Rather that solving for \(y\), we implicitly think of \(y\) as a function of \(x\), and differentiate both sides of the equation with respect to \(x\).
\[\frac{d}{dx} \left (x^2 + 6xy + y^2\right )= \frac{d}{dx}(-4)\underset{(*)}{\Longrightarrow} 2x + y + xy’ + 2y y’ =0.\]
\((*)\) Here, we have used the product rule, and have taken into account that, in general,
\[\frac{d}{dx} y^n = \frac{d}{dx} (y(x))^n = n( y(x))^{n-1} y'(x) = ny^{n-1}y’.\]
Solving for \(y’\) yields
\[y’ = -\frac{2x+y}{x+2y}.\]
Finally, substituting the coordinates of \(P(-1,1)\) into this equation for \(y’\)
\[y'(-1) = -\frac{-2-1}{-1+2}= 1.\]
As in (ii).

Clearly, implicit differentiation, constitutes a very useful procedure to find \(y’\) in this context, since it is not necessary to solve for \(y\) first. The following example makes this point even more apparent.

Example 2. (Implicit Differentiation Method) The graph of the equation \[(x^2+y^2)^2=8(x^2-y^2)\] is called a lemniscate of Bernoulli. Its graph appears in Figure 4.3. (a) Verify that \(\displaystyle P\left (-\frac{12}{5}, \frac{4}{5}\right )\) is on the graph of the lemniscate; (b) find the slope-intercept equation of the line tangent to the graph at \(P\).

Figure 4.3: Example 2

Solution. (a) Substituting the coordinates of the point in the equation, we have
\[\text{LHS} = \left (\left (-\frac{12}{5}\right )^2+\left ( \frac{4}{5} \right )^2\right )^2 = \left (\frac{160}{25}\right )^2 = \left (\frac{32}{5}\right )^2= 8\left (\frac{128}{25}\right ),\]
and
\[\text{RHS} = 8\left ( \left (-\frac{12}{5}\right )^2-\left (\frac{4}{5}\right )^2\right ) = 8\left ( \frac{128}{25}\right ).\]
Thus, the coordinates of \(P\) satisfy the equation of the lemniscate, that is, the point is on its graph.

(b) We now proceed to compute the derivative –with respect to \(x\)– on both sides of the equation keeping in mind that \(\displaystyle y’ = \frac{dy}{dx}\), and using the chain rule.

\[\frac{d}{dx}(x^2+y^2)^2=\frac{d}{dx}8(x^2-y^2) \Longrightarrow 2(x^2+y^2)(2x + 2y y’)=8(2x -2y y’)\Longleftrightarrow\]
\[\Longleftrightarrow (x^2+y^2)(x + y y’)=4(x -y y’)\]
Given that we only need to find \(\displaystyle y’\left (-\frac{12}{5}\right )\), we can directly substitute the coordinates of \(P\) into this equation. Namely
\[ \left( \left (-\frac{12}{5}\right )^2+\left (\frac{4}{5}\right )^2\right )
\left(-\frac{12}{5}+ \frac{4}{5} y’\right )
=4 \left(-\frac{12}{5} – \frac{4}{5} y’\right ) \underset{?}{\Longleftrightarrow }
2\left(-12+ 4 y’\right ) =-5 \left(3 + y’\right )\Longleftrightarrow
\]
\[\Longleftrightarrow 13y’=9\Longleftrightarrow y’ = \frac{9}{13}. \]
It follows that the slope-intercept equation of the tangent line to the lemniscate at \(P\) is given by \[\displaystyle y = \frac{9}{13}x+\frac{32}{13}\]
(verify this).

Example 3. The ellipse \( \displaystyle x^2 + \frac{y^2}{4 } = 1\) has two points whose tangent lines pass through the point \(A (3,2)\). One point is \((0,2)\) (see Figure 4.4). Find the coordinates of the other point.

Figure 4.4: Example 3

Let \(P(x,y)\) be the second point whose tangent line passes through \(A\). Since there are two unknowns, namely, the two coordinates of \(P\), it is necessary to set up two equations.

One equation is readily available: the implicit equation of the ellipse, \(\displaystyle x^2 + \frac{y^2}{4 } = 1\), as the point \(P(x,y)\) lies on the ellipse.

The key idea to set up a second equation is to \textit{compute the slope of the tangent line} passing through \(A\) and \(P\) \textit{in two different ways}.

Note that if \(y=0\), then \(x=\pm 1\), and \(y’\) is undefined, as the tangent lines to the ellipse at \((0,\pm 1)\) are vertical.

Hence, the second equation is given by \(\displaystyle \frac{y-2}{x-3}=-\frac{4x}{y}\).\\

Thus, the equations to solve simultaneously are
\[x^2 +\frac{y^2}{4}=1 \text{ and } \frac{y-2}{x-3}=-\frac{4x}{y}.\]
Given that \(P(x,y)\) is on the ellipse, \(-1\leq x \leq 1\) (explain why), thus, in particular, \(x-3\neq 0\). Furthermore, in (ii) above we pointed out that \(y\neq 0\), else \(y’\) is undefined. Hence, the second equation can be rewritten as follows:
\[\frac{y-2}{x-3}=-\frac{4x}{y} \Longleftrightarrow y(y-2) = -4x(x-3) \Longleftrightarrow 4x^2+y^2 = 12x+2y.\]
Thus, we need to solve
\[\hspace{1in}\hphantom{(*)} 4x^2+y^2=4 \text{ and } 4x^2+y^2 = 12x+2y. \hspace{1in}(*)\]
It is immediate that
\[\hspace{1.4in}\hphantom{(**)}12x+2y= 4 \Longleftrightarrow y = 2-6x.\hspace{1.4in}(**)\]
Substituting this into the first equation in \((*)\) (which is simpler than the second equation), we have
\[4x^2+\left (2-6x\right )^2=4 \Longleftrightarrow x^2 + 1 – 6x + 9x^2 = 1 \Longleftrightarrow x( 5 x -3) = 0 \Longleftrightarrow x = 0 \text{ or } x= \frac{3}{5}.\]
Substituting these values of \(x\) into \((**)\), we obtain
\[\text{if} \; x=0, \; y = 2, \text{ and } \text{ if } x= \frac{3}{5}, \; y = 2-\frac{18}{5} = -\frac{8}{5}.\]
It follows that the second point on the ellipse whose tangent line passes through \(A\) is \(\displaystyle P\left (\frac{3}{5}, -\frac{8}{5}\right )\).