From the given equation.
y=[(2x+3)/(x-2)][(5x-1)/(3x-2)]
Simplify the right side of the equation by multiplying the rational expressions first
y=[(2x+3)/(x-2)][(5x-1)/(3x-2)]
y=[(10x^2+13x-3)/(3x^2-8x+4)]
Find the derivative by using the formula d/dx(u/v)=(v*d/dx(u)-u*d/dx(v))/v^2
Let u=10x^2+13x-3
Let v=3x^2-8x+4
Use now the formula
d/dx(u/v)=(v*d/dx(u)-u*d/dx(v))/v^2
y'=d/dx((10x^2+13x-3)/(3x^2-8x+4))
y'=((3x^2-8x+4)*d/dx(10x^2+13x-3)-(10x^2+13x-3)*d/dx(3x^2-8x+4))/(3x^2-8x+4)^2
y'=((3x^2-8x+4)*(20x+13)-(10x^2+13x-3)(6x-8))/(3x^2-8x+4)^2
Simplify the numerator
y'=(60x^3-160x^2+80x+39x^2-104x+52-(60x^3+78x^2-18x-80x^2-104x+24))/(3x^2-8x+4)^2
y'=(60x^3-160x^2+80x+39x^2-104x+52-60x^3-78x^2+18x+80x^2+104x-24)/(3x^2-8x+4)^2
y'=(-119x^2+98x+28)/(3x^2-8x+4)^2
God bless....I hope the explanation is useful.
