Differentiate: y to y'?

Question

Differentiate y = (4x - 7) / (2 x + 3)


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Thank you for helping!!

Answer 1

You need to use the quotient rule, where (f/g)' = (gf' - fg')/(g^2).

So y = (4x - 7) / (2x + 3), and you should be able to calculate that the derivative of 4x - 7 is 4 and the derivative of 2x + 3 is 2. Therefore y' = ((2x + 3)4 - (4x - 7)2) / (2x + 3)^2 = (8x + 12 - (8x - 14)) / (2x + 3)^2 = -2 / (4x^2 + 12x + 9).

Answer 2

Ok you have two choices, you can use the product rule or the quotient rule. I used the quotient rule.

Quotient Rule: (g(x)f'(x) - f(x)g'(x))/ g(x)^2

Where f(x) = (4x - 7) and g(x) = (2x + 3)

f'(x) = 4 and g'(x) = 2 so the equation should look like this:

((2x +3)(4) - (4x - 7)(2))/ (2x+3)^2

8x +12 - (8x -14) / (2x+3)^2

distribute the negative sign then combine like terms to get,

26/ (2x +3)^2

and you can expand (2x +3)^2 to get 4x^2 +12x +9
and your final answer would be

26/ 4x^2 +12x +9

Hope this helped

Answer 3

To differentiate this you need to follow the quotient rule, where y = f(x)/g(x) and y' = [g(x).f'(x) - f(x).g'(x)] / [g(x)]^2
So y' = [(2x + 3)(4) - (4x - 7)(2)] / (2x + 3)^2
y' = (8x + 12 - 8x + 14) / (2x + 3)^2
y' = (24) / (2x + 3)^2