Chain rule

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# Chain rule
## Differential Calculus
### Arturo Sánchez González
### <a href="mailto:arturo.sanchez@upaep.mx" class="email">arturo.sanchez@upaep.mx</a>
### May 2022

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# Chain rule

Let `$\color{red}{g(x)}$` and `$\color{blue}{f(x)}$` be two functions. Then `$\color{green}{h(x)} = \color{red}{g}\left( \color{blue}{f(x)} \right)$` is also a function and

`$$\color{darkgreen}{h'(x)} = \color{darkred}{g'}\left( \color{blue}{f(x)}\right) \cdot \color{darkblue}{f'(x)}$$`

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# How to apply the chain rule ? (1/3)

Consider the function `$\color{green}{h(x) = e^{x^2 - 1}}$`.

- **1st step: Identify the inner and the outer functions**

In this case, let's consider `$\color{red}{g(x) = e^{x}}$` and `$\color{blue}{f(x) = x^2-1}$`.

**How to note this?** Just answer the following question:

<div align='center', style="font-size:25px"> 
 Where do I have to make the first substitution? 
</div>

The answer to the previous question gives us the inner function . In this example is `$\color{blue}{f(x) = x^2 - 1}$`.

To get the outer function we just have to observe the next expression that we need to apply. In this example, after calculating `$\color{blue}{f(x)}$` we have to take the exponential of that value, then `$\color{red}{g(x) = e^{x}}$`. Observe that

`$$\color{red}{g}(\color{blue}{f(x)}) = \color{red}{g}(\color{blue}{x^{2} - 1} ) = \color{red}{e}^{\color{blue}{x^{2}-1}} = \color{green}{h(x)}$$`

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# How to apply the chain rule ? (2/3)

- 2nd step: Find the derivatives of the inner and the outer functions

Here, we have that

`$$\begin{aligned}
\color{darkred}{g'(x)} & = \color{darkred}{e^{x}} \\
\color{darkblue}{f'(x)} & = \color{darkblue}{2x}
\end{aligned}$$`

<hr>

- 3rd step: Evaluate the derivative of the outer function in the value of the inner function

This step means to find `$\color{darkred}{g'}\left( \color{blue}{f(x)} \right)$`. Then, we have that

`$$\color{darkred}{g'}\left(\color{blue}{f(x)}\right) = \color{darkred}{g'}(\color{blue}{x^2 -1}) = \color{darkred}{e}^{\color{blue}{x^2-1}}$$`

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# How to apply the chain rule ? (3/3)

- 4th step: Multiply the value obtained in the previous step by the derivative of the inner function

We already know that `$\color{darkred}{g'} \left( \color{blue}{f(x)} \right) = \color{darkred}{e}^{\color{blue}{x^2-1}}$` and `$\color{darkblue}{f'(x)} = \color{darkblue}{2x}$`, then

`$$\color{darkgreen}{h'(x)} = \color{darkred}{g'} \left( \color{blue}{f(x)} \right) \cdot \color{darkblue}{f'(x)} = \color{darkred}{e}^{\color{blue}{x^2-1}} \left( \color{darkblue}{2x} \right)$$`

<hr>

- 5th step: Make all the necessary reductions

In our example we only have to reorder the terms, but in practice is usually necessary to make more calculations. Then, our final answer is

`$$\color{darkgreen}{h'(x)} = \color{darkblue}{2x} \color{darkred}{e}^{\color{blue}{x^2-1}}$$`

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# Examples (1/2)
 Find the derivative of the following function.

`$$\color{darkgreen}{\ell(t) = \left( \ln(t) \right)^2 }$$`

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# Examples (2/2)

Find the derivative of the following function.

`$$\color{darkblue}{r(t) = \ln \left( t^2\right) }$$`

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# Dig deeper

Find the derivative of the following function.

`$$\color{darkred}{m(x) = 7\cos\left( e^{x} - 2x^5 \right)}$$`