Chain rule

Function Composition

Before we move on to Chain rule, we discuss function composition. Suppose we have two functions $f(x)$ and $g(x)$; the composition of $f(x)$ and $g(x)$, denoted by $(f \circ g)(x)$, is evaluated by plugging the second function in the first function, as follows

Definition

Suppose $f(u)$ is differentiable at $u=g(x)$ and $g(x)$ is differentiable at $x$. It follows that the composition of $f$ and $g$, i.e. $(f \circ g)(x)=f(g(x))$ is differentiable at $x$ and that

Letting $y=f(u)$ and $u=g(x)$, then

where $d y / d u$ is evaluated at $u=g(x)$.

Proof of chain rule

From first principles, $f(g(x))^{\prime}$ is

The answer involves the derivative of the outer function, $f$ and of the inner function, $g$. Assuming $g^{\prime}$ exists ( $g$ is differentiable at $x$ ), using the definition of the derivative, we have

as $\Delta x \rightarrow 0$. We introduce a new variable $v$ to define this quantity as follows,

where $v \rightarrow 0$ as $\Delta x \rightarrow 0$.

We rearrange to give:

Next, we assume that $f$ is differentiable at $u=g(x)$ and define a new variable $w$ (similar to $v$ above) giving

note that this is a result of the limit definition of $f^{\prime}(u)$ and $w \rightarrow 0$ as $\Delta u \rightarrow 0$.

We rearrange to give:

Now, substituting the first rearranged eqn in $f(g(x+\Delta x))$ gives

Comparing the RHS of this eqn and the LHS of the second rearranged eqn, we have that $\Delta u=\Delta x\left[v+g^{\prime}(x)\right]$ and, of course, $u=g(x)$. It follows that, $\Delta u \rightarrow 0$ as $\Delta x \rightarrow 0$ and so $w \rightarrow 0$ as $\Delta x \rightarrow 0$ (we will make use of these in the last step of the proof). Using the above expressions for $\Delta u$ and $u$ in the second rearranged eqn, we have

Subsituting these eqns back into one another, we get:

which simplifies to

Using the sum and product limit properties, we finally have

since $v \rightarrow 0, w \rightarrow 0$ as $\Delta x \rightarrow 0$.

Applications of the chain rule

The chain rule is therefore used to differentiate composite functions. For example consider $f(x)=e^{x}$ and $g(x)=\sin x$. Then, $(f \circ g)(x)=\exp (\sin x)$ and its derivative is

Another application of the chain rule is implicit differentiation. For example, compute the slope of the tangent to the unit circle $x^{2}+y^{2}=1$. To differentiate explicitly, we first need to express $y=f(x)$ as follows

whose derivative is

To differentiate implicitly, we differentiate $x^{2}+y^{2}=1$ wrt $x$, treating $y$ as a function of $x$ and $y^{2}$ as a composite function of $x$ so that

Therefore by implicit differentiation we have

This is equivalent to the result obtained with explicit differentiation. It is not always easy (or possible) to express $y$ as an explicit function of $x$ in order to use explicit differentiation. With implicit differentiation, we can still obtain an expression for the derivative and evaluate slopes at given coordinates $(x, y)$.

For another example, if we wanted to compute the derivatives of the curve $x^{2}=y^{3}-y$ using implicit differentiation and evaluate the slope at $(\sqrt{6}, 2)$.

Differentiating implicitly we have

or

At $(\sqrt{6}, 2)$, the slope is

Further, implicit differentiation is used for logarithmic differentiation. Keeping in mind that $\log _{a} x$ and $a^{x}$ are inverse functions $(a>0, x>0, a \neq 1)$, consider a function $x=a^{y}$ or, equivalently, $y=\log _{a} x$. Now, we have

which we can show is true by taking the natural logarithm on both sides. Then,

which gives

Solving for $y$, we have

but also, $y=\log _{a} x$ yielding the relationship,

From Eq. (3.17), we can differentiate logarithms with any base: