Differentiation is the process through which we can find the rate of change of a dependent variable in relation to a change of the independent variable. It is also called a derivative. We have to have at least one variable which we consider the Independent Variable and a second variable as the Dependent Variable which is related to the independent variable in some way. Sometimes for the complex and mixed type of functions, finding the derivative is very hard. For such problems, the chain rule is very effective. In this topic, we will discuss the chain rule formula. Let us learn it!

**What is the Chain Rule?**

The chain rule provides us a technique for determining the derivative of composite functions. It is applicable to the number of functions that make up the composition. Therefore, the chain rule is providing the formula to calculate the derivative of a composition of functions. Let us suppose that f and g are the functions, then the chain rule will express the derivative of their composition.

Source:en.wikipedia.org

**Chain Rule Formula**

The Chain Rule Formula is as follows –

Let us suppose,

y = f(x) i.e. y is a function of x and y = f(u) i.e. y is a function of u

u = f (x) i.e. u is a function of x

Then** \(\frac{dy}{dx}\)= \(\frac{dy}{du}\times\frac{du}{dx}\)**

For example to find differentiation of following function,

\(y = cos x^2\)

We will apply the chain rule as follows.

\(y = cos x^2\) Let \(u = x^2\), then we have \(y = cos u\)

Therefore: \(\frac{du}{dx} = 2x\)

and \(\frac{dy}{du} = -sin u\)

and so, the chain rule says:

\(\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}\)

\(\frac{dy}{dx} = -sin u \times 2x\)

\(\frac{dy}{dx} = – 2x sin x^2\)

Thus the derivative of y with respect to x will be \((- 2x sin x^2).\)

Thus we can see that this method of chain rule will sometime make the difficult process of differentiation as a simple computation.

**Solved Examples for Chain Rule Formula**

Q.1: Let f(x) = 6x + 3 and g(x) = -2x+5 . Using the chain rule determine h'(x) where h(x) = f(g(x)).

Solution: The derivatives of f and g are:

\(\begin{align*} f'(x)&=6 g'(x)&=-2 \end{align*}\)

According to the chain rule,

\(\begin{align*} h'(x) &= f'(g(x)) g'(x)\\ &= f'(-2x +5) (-2)\\&= 6 (-2)=-12. \end{align*}\)

Since both the functions were linear, so it was trivial. Also, we had to evaluate f’ at g(x) = -2x+5, which didn’t make a difference, because f’ = 6 not matter what its input is. Thus, in this case, if we calculate h(x),

\(\begin{align*} h(x) &= f(g(x))\\&= f(-2x+5)\\ &= 6(-2x+5)+3\\ = -12x+30+3 = -12x + 33, \end{align*}\)

Then we can easily calculate its derivative directly and finally we can obtain that h'(x) = -12.

Q.2: Let \(f(x)=e^x and g(x)=4x\) . Use the chain rule to calculate h'(x) where h(x)=f(g(x)).

Solution: The derivative of the exponential function with base e is just the function itself, so |(f'(x) = e^x\)

The derivative of g is g'(x)=4

According to the chain rule,

\(\begin{align*} h'(x) &= f'(g(x)) g'(x)\\ &= f'(4x) \cdot 4\\ &= 4e^{4x}. \end{align*}\)

In this example, it was important that we had to evaluate the derivative of f at 4x. Then the derivative of \(h(x) = f(g(x)) = e^{4x}\) is not equal to \(4e^{x}\)

The correct answer is \(h'(x) = 4e^{4x}.\)

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