Question

Find `dy/dx, "if"  y=sqrt((2x+3)^5/((3x-1)^3(5x-2)))`

Answer 1

`y=sqrt((2x+3)^5/((3x-1)^3(5x-2)))`

∴ `logy = log[(2x+3^5)/((3x-1)^3(5x-2))]^(1/2)`

`=1/2log[(2x+3^5)/((3x-1)^3(5x-2))]`

`=1/2[log(2x+3)^5-log(3x-1)^3-log(5x-2)]`

`=1/2[5log(2x+3)-3log(3x-1)-log(5x-2)]`

`=5/2log(2x+3)-3/2log(3x-1)-1/2log(5x-2)`

Differentiating both sides w.r.t. x, we get

`1/y.dy/dx=5/2d/dx[log(2x+3)]-3/2d/dx[log(3x-1)]-1/2d/dx[log(5x-2)]`

`=5/2xx1/(2x+3)d/dx(2x+3)-3/2xx1/(3x-1)d/dx(3x-1)-1/2xx1/(5x-2)d/dx(5x-2)`

`=5/(2(2x+3))xx(2xx1+0)-3/(2(3x-1))xx(3xx1-0)-1/(2(5x-2))xx(5xx1-0)`

∴`dy/dx=y[5/(2x+3)-9/(2(3x-1))-5/(2(5x-2))]`

`=sqrt((2x+3)^5/((3x-1)^3(5x-2)))[5/(2x+3)-9/(2(3x-1))-5/(2(5x-2))]`