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