# Chain Rule

In this method chain rule we are going to different problems to understand where we have to apply this method.

Example 1:

Differentiate log √x

Solution:

Let y = log √x

Here we have the function log √x. In this we have two function which are interrelated. So let us take t = √x. Now the function is going to become y = log t.

Formula:

dy/dx = (dy/dt) x (dt/dx)

t = √x                                        y = log t

dt/dx = (x)^1/2                        dy/dt = 1/t

dt/dx = (1/2) x^(1/2-1)

= (1/2)x^-1/2

= 1/2√x

dy/dx = (1/t) 1/2√x

= 1/2√x t

= 1/2√x √x

= 1/2x

Example 2:

Differentiate log sin (x² + 2x + 3)

Solution:

Let y = log sin (x² + 2x + 3)

Here three functions are related.

(i) Take t = (x² + 2x + 3).

(ii) Now the function is becoming y = log sin t.

(iii) Now take u = sin t

(iv) Now the function is becoming y = log u

Formula

dy/dx = (dy/du) x (du/dt) x (dt/dx)

y = log u             u = sin t               t = x² + 2x + 3

dy/du = 1/u     du/dt = cos t        dt/dx = 2x + 2 (1) + 0

dt/dx = 2x + 2

dy/dx = (1/u) ( cos t ) (2x + 2)

= cos t/u  (2x + 2)

= cos t/sin t  (2x + 2)

= cot t (2x+2)

du/dx = (2x+2) cot (x² + 2x + 3)

Example 3:

Differentiate log (tan x)

Let y = log (tan x)

Take t = tan x. So the function y is going to become y = log t.

Formula:

dy/dx = (dy/dt) x  (dt/dx)

y = log t                t = tan x

dy/dt = 1/t            dt/dx = sec² x

dy/dx = (dy/dt) x (dt/dx)

=  (1/t) x sec² x

= sec² x/t

= sec² x/tan x

= (1/ cos ² x) /(sin x/cos x )

= (1/cos ²x) x(cos x /sin x )

= sinx /cos x

= tan x

Chain rule to First principles