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Section The Chain Rule

We can take the derivative of \(\dsp f(x) = \sqrt{x}\) and \(g(x) = 5x+1\) easily, but to take the derivative of \(\dsp h(x) = \sqrt{5x+1},\) we would need to return to the limit definition for the derivative. We can take the derivative of \(f(x) = \sin(x)\) and g(x) = \(x^2\) easily, but to take the derivative of \(h(x) = \sin(x^2)\) we would again return to the limit definition. The chain rule remedies this problem by giving a rule for computing the derivatives of functions that are the composition of two functions.

Algebra Reminder. The function \(h(x) = \sqrt{5x+1}\) is the composition of two functions \(f(x) = \sqrt{x}\) and \(g(x) = 5x + 1\) since we may write \(f(g(x)) = f(5x +1) = \sqrt{5x+1} = h(x).\) The function \(h(x) = \sin(x^2)\) is the composition of \(f(x) = \sin(x)\) and \(g(x) = x^2.\) If \(h\) equals \(f\) composed with \(g\text{,}\) we write \(h = f \circ g\) and \(h(x) = f(g(x))\) for all numbers \(x\) in the domain of \(h.\)

A sort-of proof. Assume that each of \(f\) and \(g\) is a differentiable function and that \(h(x) = f(g(x))\text{.}\) From the limit definition of the derivative of \(h\text{,}\) we know that

\begin{equation*} h'(x) = \lim_{t \rightarrow x} \frac{h(t)-h(x)}{t-x}. \end{equation*}

Since \(h(x) = f(g(x))\text{,}\)

\begin{equation*} h'(x) = \lim_{t \rightarrow x} \frac{f(g(t))-f(g(x))}{t-x}. \end{equation*}

Since

\begin{equation*} \frac{g(t)-g(x)}{g(t)-g(x)} = 1 \mbox{ as long as g(t) } \neq \mbox{ g(x) } , \end{equation*}

by using the Product Rule for Limits we may write,

\begin{equation*} h'(x) = \lim_{t \rightarrow x} \frac{f(g(t))-f(g(x))}{g(t)-g(x)} \cdot \frac{g(t)-g(x)}{t-x} = \lim_{t \rightarrow x} \frac{f(g(t))-f(g(x))}{g(t)-g(x)} \cdot \lim_{t \rightarrow x} \frac{g(t)-g(x)}{t-x}. \end{equation*}

Since \(g\) is differentiable, we know by Theorem 2.24 that \(g\) is continuous, so by the definition of continuity, we have \(\dsp \lim_{t \rightarrow x} g(t)=g(x).\) Thus, in the first limit, we can replace \(t \rightarrow x\) with \(g(t) \rightarrow g(x).\)

\begin{equation*} h'(x) = \lim_{g(t) \rightarrow g(x)} \frac{f(g(x))-f(g(t))}{g(x)-g(t)} \cdot \lim_{t \rightarrow x} \frac{g(x)-g(t)}{t-x}. \end{equation*}

The right-hand side of this equation is the product of two limits. The second limit is the definition of \(g'(x)\text{.}\) Substituting \(A = g(t)\) and \(B = g(x)\) into the first limit, we may write this as,

\begin{equation*} \lim_{A \rightarrow B} \frac{f(B)-f(A)}{B-A}. \end{equation*}

But this is just \(f'(B)\text{,}\) or \(f'(g(x))\text{,}\) so we have shown that

\begin{equation*} h'(x) = f'(g(x))\cdot g'(x). \end{equation*}

q.e.d.

Problem 2.53.

Let \(h(x) = \sqrt{7+\sin(x)}.\)

  1. Find functions f and g such that \(h = f \circ g.\)

  2. Compute \(f'.\)

  3. Compute \(f' \circ g.\)

  4. Compute \(g'.\)

  5. Use the Chain Rule to compute \(h'.\)

Problem 2.54.

Let \(h(x) = \cos(x^2)\text{.}\)

  1. Find functions f and g such that \(h = f \circ g.\)

  2. Compute \(f'.\)

  3. Compute \(f' \circ g.\)

  4. Compute \(g'\text{.}\)

  5. Use the Chain Rule to compute \(h'\text{.}\)

Problem 2.55.

Given that \(f(x) = (x^{5/2} - 4x^{1/3} + 365 )^{42}\text{,}\) compute \(f'.\)

Problem 2.56.

If \(y = (\cos(x^2) )^2\text{,}\) compute \(y'\text{.}\)

Problem 2.57.

Let \(\dsp y = \left( \frac{1-x^2}{1+x^2} \right)^{10}\) and compute \(y'\) using the Chain Rule first and then the Quotient Rule. Check your answer by rewriting \(\dsp y = \frac{(1-x^2)^{10}}{(1+x^2)^{10}}\) and computing \(y'\) using the Quotient Rule first and then the Chain Rule.

Problem 2.58.

Find a function f with derivative \(f'(x) = 5x + 3.\)

Problem 2.59.

Find the equation of the tangent line to the curve \(y = (x + 1/x)^3\) at the point where \(x = -1\text{.}\) Graph the curve and the line.

Problem 2.60.

Assume \(a, b, c\) and \(d\) are real numbers and \(f(w) = a (\cos(wb) )^2 + c(\sin(wd) )^2.\) Compute \(f'\text{.}\)

Problem 2.61.

Find the real number m such that \(y = m \cos(2t)\) satisfies the differential equation \(y'' + 5y = 3\cos(2t).\)

Problem 2.62.

Given that \(f'(x) = \sqrt{2x+3},\) \(g(x) = x^2 + 2\text{,}\) and \(F(x) = f(g(x))\text{,}\) compute \(F'.\)

Problem 2.63.

Given that \(\dsp f'(x) = \frac{x}{x^2-1}\) and \(g(x) = \sqrt{2x-1}\text{,}\) compute \(F'\) where \(F(x) = f(g(x))\text{.}\)

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