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Question
Find the derivative of the following w. r. t.x.
`(3x^2 - 5)/(2x^3 - 4)`
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Solution
Let y = `(3x^2 - 5)/(2x^3 - 4)`
using `dy/dx = (v(du)/dx - u(dv)/dx)/v^2`
let `(3x^2 - 5) = u and (2x^3 - 4) = v`
Differentiating w.r.t. x, we get
`dy/dx=d/dx((3x^2 - 5)/(2x^3 - 4))`
= `((2x^3 - 4)d/dx(3x^2 - 5) - (3x^2 - 5)d/dx(2x^3 - 4))/(2x^3 - 4)^2`
= `((2x^3 - 4)(3d/dxx^2 - d/dx5) - (3x^2 - 5)(2d/dxx^3 - d/dx4))/(2x^3 - 4)^2`
= `((2x^3 - 4)[3(2x - 0)] - (3x^2 - 5)[2(3x^2 - 0)])/(2x^3 - 4)^2`
= `((2x^3 - 4)[3(2x)] - (3x^2 - 5)[6x^2])/(2x^3 - 4)^2`
= `((2x^3 - 4)(6x) - (3x^2 - 5)(6x^2))/(2x^3 - 4)^2`
= `(6x(2x^3 - 4) - 6x^2(3x^2 - 5))/(2x^3 - 4)^2`
= `(12x^4 - 24x - 18x^4 + 30x^2)/(2x^3 - 4)^2`
= `(-6x^4 + 30x^2 - 24x)/(2x^3 - 4)^2`
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