Chain Rule in Differentiation with Examples

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