Here you will learn what is chain rule in differentiation with examples.

Let’s begin –

## Chain Rule in Differentiation

If f(x) and g(x) are differentiable functions, then fog is also differentiable and

(fog)'(x) = f'(x) = f'(g(x)) g'(x)

or, \(d\over dx\) {(fog) (x)} = \(d\over d g(x)\) {(fog) (x)} \(d\over dx\) (g(x)).

**Remark 1 **: The above rule can also be restated as follows :

If z = f(y) and y = g(x), then \(dz\over dx\) = \(dz\over dy\).\(dy\over dx\)

Derivative of z with respect to x = (Derivative of z with respect to y) \(\times\) (Derivative of y with respect to x)

**Remark 2** : This chain rule can be extended further.

Derivative of z with respect to x = (Derivative of z with respect to u) \(\times\) (Derivative of u with respect to v) \(\times\) (Derivatve of v with respect to x)

**Example 1** : Differentiate \(sin(x^2 + 1)\) with respect to x.

**Solution** : Let y = \(sin(x^2 + 1)\). Putting u = \(x^2 + 1\) , we get

y = sin u and u = \(x^2 + 1\)

\(\therefore\) \(dy\over du\) = cosu and \(du\over dx\) = 2x

Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\)

\(\implies\) \(dy\over dx\) = (cos u)2x = 2x \(cos (x^2 + 1)\)

Hence, \(d\over dx\) [\(sin(x^2+1)\)] = 2x \(cos (x^2 + 1)\)

**Example 2** : Differentiate \(e^{sinx}\) with respect to x.

**Solution** : Let y = \(e^{sinx}\). Putting u = sinx , we get

y = \(e^u\) and u = sinx

\(\therefore\) \(dy\over du\) = \(e^u\) and \(du\over dx\) = cosx

Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\)

\(\implies\) \(dy\over dx\) = \(e^u\)cosx = \(e^{sinx}\)cosx

Hence, \(d\over dx\) [\(e^{sinx}\)] = \(e^{sinx}\)cosx

**Example 3** : Differentiate log sinx with respect to x.

**Solution** : Let y = log u. Putting u = sinx , we get

y = log u and u = sinx

\(\therefore\) \(dy\over du\) = \(1\over u\) and \(du\over dx\) = cosx

Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\)

\(\implies\) \(dy\over dx\) = \(1\over u\)cosx = \(1\over sinx\)cosx = cotx

Hence, \(d\over dx\) [log sinx] = cotx