# Chain rule in calculus

## Statement

The derivative of a nested function is the derivative of the outer function multiplied by the derivative of the inner function.

$$\tfrac{d}{dx}\bigg(g(j(x))\bigg) = g'(j(x)) * j'(x)$$

## Proof

Define function $f$.

$$f(x) = g(j(x))$$

$$f'(x) = \lim_{h \to 0}\left(\frac{f(x + h) - f(x)}{h}\right)$$

$$f'(x) = \lim_{h \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{h}\right)$$

Multiply the numerator and denominator of the fraction by $j(x + h) - j(x)$ and split the limit.

$$f'(x) = \lim_{h \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{h} * \frac{j(x + h) - j(x)}{j(x + h) - j(x)}\right)$$

$$f'(x) = \lim_{h \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{j(x + h) - j(x)} * \frac{j(x + h) - j(x)}{h}\right)$$

$$f'(x) = \lim_{h \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{j(x + h) - j(x)}\right) * \lim_{h \to 0}\left(\frac{j(x + h) - j(x)}{h}\right)$$

Let $k = j(x + h) - j(x)$. From this follows that $j(x + h) = k + j(x)$.

Now we can conclude that as $h$ approaches to $0$, so does $k$:

$$k = j(x + h) - j(x) = j(x + 0) - j(x) = j(x) - j(x) = 0$$

Substitute $k$ in the derivative.

$$f'(x) = \lim_{k \to 0}\left(\frac{g(j(x + h)) - g(j(x))}{k}\right) * \lim_{h \to 0}\left(\frac{j(x + h) - j(x)}{h}\right)$$

Here are the definitions of the derivative for $g(j(x))$ and $j(x)$.

$$f'(x) = g'(j(x)) * j'(x)$$

## Proofs building upon this proof

### Exponent rule in calculus

This proofs show the derivative of a^x is a^x * ln(a).

### Power rule in calculus

This proofs shows that the derivative of x^r is r * x^(r - 1).

### Quotient rule in calculus

This proof shows that the derivative for the quotient or fraction a/b is (a'b - ab') / b^2.

### The derivative of ln(x)

This proof shows that the derivative of ln(x) is 1/x.