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# 6. Chain Rule

### Chain Rule

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 $f'(b)g'(a) = f'(g(a))g'(a)$

### Examples

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 $-e^x\text{sin}(e^x)$