Differentiation of a function of a function

It sometimes happens that $y$, instead of being defined directly as a function of $x$, is given as a function of another variable $v$, which is defined as a function of $x$. In that case $y$ is a function of $x$ through $v$ and is called a function of a function or a composite function. The process of substituting one function into another is sometimes called composition.

For example, if $y = \frac{2v}{1 - v^2}$, and $v = 1-x^2$, then $y$ is a function of a function. By eliminating $v$ we may express $y$ directly as a function of $x$, but in general this is not the best plan when we wish to find $\frac{dy}{dx}$.

If $y = f(v)$ and $v = g (x)$, then $y$ is a function of $x$ through $v$. Hence, when we let $x$ take on an increment $\Delta x$, $v$ will take on an increment $\Delta v$ and $y$ will also take on a corresponding increment $\Delta y$. Keeping this in mind, let us apply the General Rule simultaneously to the two functions $y = f(v)$ and $v = g (x)$.

• FIRST STEP. $y + \Delta y = f(v + \Delta v)$, $v + \Delta v = g(x + \Delta x)$.

• SECOND STEP.
  $$\begin{array}{rlcrl} y + \Delta y & = f(v + \Delta v), & \qq... ...(v), & \qquad & \Delta v &=g (x + \Delta x) - g (x) \end{array}$$

• THIRD STEP. $\frac{\Delta y}{\Delta v} = \frac{f(v + \Delta v) - f(v)}{\Delta v}$, $\frac{\Delta v}{\Delta x} = \frac{g(x + \Delta x) - g(x)}{\Delta x}$.

  The left-hand members show one form of the ratio of the increment of each function to the increment of the corresponding variable, and the right-hand members exhibit the same ratios in another form. Before passing to the limit let us form a product of these two ratios, choosing the left-hand forms for this purpose.

  This gives $\frac{\Delta y}{\Delta v} \cdot \frac{\Delta v}{\Delta x}$, which equals $\frac{\Delta y}{\Delta x}$. Write this

  $$\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta v} \cdot \frac{\Delta v}{\Delta x}.$$

• FOURTH STEP. Passing to the limit,

  $$\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx},$$ (5.1)

  
  
  by Theorem 3.8.2.This may also be written
  $$\frac{dy}{dx} = f'(v) \cdot g'(x).$$

The above formula is sometimes referred to as the chain rule for differentiation. If $y = f(v)$ and $v = g (x)$, the derivative of $y$ with respect to $x$ equals the product of the derivative of $y$ with respect to $v$ and the derivative of $v$ with respect to $x$.