Question

If `f(x)=(2x+3)/(3-2x)`, differentiate `y=sin(logf(x))` using chain rule?

Answer 1

`(dy)/(dx)=12/(9-4x^2)cos(log((2x+3)/(3-2x)))`

Explanation:

As `y=f((2x+3)/(3-2x))`, this means `y=sin(log((2x+3)/(3-2x)))`

and using function of a function, we have

`(dy)/(dx)=cos(log((2x+3)/(3-2x)))xx1/((2x+3)/(3-2x))xx(2(3-2x)-(-2)(2x+3))/(3-2x)^2`

= `cos(log((2x+3)/(3-2x)))xx(3-2x)/(2x+3)xx(6-4x+4x+6)/(3-2x)^2`

= `cos(log((2x+3)/(3-2x)))xx12/((2x+3)(3-2x))`

= `12/(9-4x^2)cos(log((2x+3)/(3-2x)))`