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Let \(g\) be a function that is differentiable at the point \(a\) and let \(f\) be a function that is differentiable at the point \(g(a).\) The chain rule says that \[(f(g(a))' = f'(g(a))g'(a)\]

▼ Proof:

By the definition of derivative,
\begin{align}
f(g(a))' & = \lim_{x \rightarrow a} \frac{f(g(a))-f(g(x))}{a-x} \\
& = \lim_{x \rightarrow a} \frac{f(g(a))-f(g(x))}{g(a)-g(x)} \cdot \frac{g(a)-g(x)}{a-x} \\
& = \left[\lim_{x \rightarrow a} \frac{f(g(a))-f(g(x))}{g(a)-g(x)} \right]\left[\lim_{x \rightarrow a}\frac{g(a)-g(x)}{a-x} \right]
\end{align}
Since \(g\) is differential at \(a,\) it is continuous at \(a.\) Therefore, as \(x\) approaches \(a,\) \(g(x)\) approaches \(g(a).\) Therefore, in the left limit, we can replace \(g(a)\) with \(b,\) \(g(x)\) with \(y,\) and \(\lim_{x \rightarrow a}\) with \(y \rightarrow b\) to get
\begin{align}
& \left[\lim_{x \rightarrow a} \frac{f(g(a))-f(g(x))}{g(a)-g(x)} \right]\left[\lim_{x \rightarrow a}\frac{g(a)-g(x)}{a-x} \right] \\
& = \left[\lim_{y \rightarrow b} \frac{f(b)-f(y)}{b-y} \right]\left[\lim_{x \rightarrow a}\frac{g(a)-g(x)}{a-x} \right]
\end{align}
By definition, the left limit is the derivative of \(f\) at \(b.\) So, we get
\[\left[\lim_{y \rightarrow b} \frac{f(b)-f(y)}{b-y} \right]\left[\lim_{x \rightarrow a}\frac{g(a)-g(x)}{a-x} \right] = f'(b)g'(a)\]
We replaced \(g(a)\) with \(b,\) so by substituting \(g(a)\) back in for \(b\) we get
\[f'(b)g'(a) = f'(g(a))g'(a)\]

Find the derivative of \(\text{sin}(x^2).\)

▼ Solution:

First, recognize that the inner function is \(g(x) = x^2\) and the outer function is \(f(x) = \text{sin}(x).\) We can check this:
\[f(g(x)) = f(x^2) = \text{sin}(x^2)\]
The derivative of \(g(x)\) is \(g'(x) = 2x\) and the derivative of \(f(x)\) is \(f'(x) = \text{cos}(x).\) The chain rule says \(f(g(x))' = f'(g(x))g'(x).\) In this example, \(f'(g(x)) = f'(x^2) = \text{cos}(x^2).\) So, the derivative of \(\text{sin}(x^2)\) is
\[2x\text{cos}(x^2)\]

Find the derivative of \(e^{x^3}.\)

▼ Solution:

First, recognize that the inner function is \(g(x) = x^3\) and the outer function is \(f(x) = e^x.\) We can check this:
\[f(g(x)) = f(x^3) = e^{x^3}\]
The derivative of \(g(x)\) is \(g'(x) = 3x^2\) and the derivative of \(f(x)\) is \(f'(x) = e^x.\) The chain rule says \(f(g(x))' = f'(g(x))g'(x).\) In this example, \(f'(g(x)) = f'(x^3) = e^{x^3}.\) So, the derivative of \(e^{x^3}\) is
\[3x^2e^{x^3}\]

Find the derivative of \(\text{cos}(e^x).\)

▼ Solution:

First, recognize that the inner function is \(g(x) = e^x\) and the outer function is \(f(x) = \text{cos}(x).\) We can check this:
\[f(g(x)) = f(e^x) = \text{cos}(e^x)\]
The derivative of \(g(x)\) is \(g'(x) = e^x\) and the derivative of \(f(x)\) is \(f'(x) = -\text{sin}(x).\) The chain rule says \(f(g(x))' = f'(g(x))g'(x).\) In this example, \(f'(g(x)) = f'(e^x) = -\text{sin}(e^x).\) So, the derivative of \(\text{cos}(e^x)\) is
\[-e^x\text{sin}(e^x)\]