Differentiation?

Question

Differentiate In (4x + 5 / 2 - 3x) with respect to x.

Ans: 23 / (4x + 5) (2 - 3x)

Please show me your workings and explain...I need to understand...thank you

Answer 1

Let y = In[(4x + 5) / (2 - 3x)]

Use the chain rule.
Let u = (4x + 5) / (2 - 3x)
Then y = ln(u)

dy/dx = dy/du * du/dx

dy/du = 1/u = (2-3x) / (4x+5)
Use quotient rule to find du/dx:
du/dx = [(2-3x)(4) - (4x+5)(-3)] / (2-3x)^2

So
dy/dx = {(2-3x) / (4x+5)} * {[(2-3x)(4) - (4x+5)(-3)] / (2-3x)^2}
Simplify:
= [(2-3x)(4) - (4x+5)(-3)] / [(2-3x)(4x+5)]
= (8 - 12x + 12x + 15) / [(2-3x)(4x+5)]
= 23 / [(2-3x)(4x+5)]

Answer 2

i think you meant to write the equation as ln{(4x+5)/(2-3x)} so that is how im going to solve it

the derivative of ln(x)= (1/x)*(x') which is ((2-3x)/(4x+5))*the derivative of (4x+5)/(2-3x). in order to differentiate that you have to use the quotient rule so you get ((2-3x)(4)-(4x+5)(-3))/(2-3x)^2. if you simplify that you get 23/(2-3x)^2. when you multiply that by (2-3x)/(4x+5) you end up with 23 / (4x + 5) (2 - 3x)

Answer 3

y = In [(4x + 5) / (2 - 3x)]
y = ln(4x+5) -ln(2-3x)
y' = 4/(4x+5) + 3/(2-3x)
y' = [4(2-3x)+3(4x+5)]/[(4x+5)(2-3x)]
y' = (8 -12x +12x +15)/[(4x+5)(2-3x)]
y' = 23/[(4x+5)(2-3x)]