How do you find the derivative of #y= [(2x+3)/(x-2)][(5x-1)/(3x-2)]# using the chain rule?

1 Answer

Answer:

#y'=(-119x^2+98x+28)/(3x^2-8x+4)^2#

Explanation:

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.