find the derivative of (x^2+7) / (x-4)?

Answer 1

Hi Dear Bunky;

{ as you know if;
f(x) = u(x) / v(x) ;
f '(x) = u ' v - u v ' / (v) ²
It is the formula you should use.}

-Now ,
f(x) = (x ² +7) / (x-4)
let u = x ² +7 , u' = 2x
let v = x - 4 , v' = 1

-Now follow the formula;
f '(x) = u ' v - u v ' / (v) ²
f '(x) = [(2x)( x - 4) - ( x ² +7) (1) ] / (4 -x) ²
f '(x) = [- 8x + 2x ² - x ² -7 ] / (4 -x) ²
f '(x) = ( x ² - 8x - 7) / (4 -x) ²

Good Luck.

Answer 2

This can even be done simpler

(x^2+7)/(x-4) =(x+4)+23 /(x-4) using synthetic division

differentiating we get 1-23(x-4)^-2

Answer 3

f(x) = (x^2 + 7)/(x - 4)
f'(x) = [(x - 4)(2x) - (x^2 + 7)]/(x - 4)^2
= (2x^2 - 8x - x^2 + 7)/(x - 4)^2
= (x^2 - 8x - 7)/(x - 4)^2

Answer 4

f(x) = (x^2 + 7)/(x - 4)

f'(x) = (((x - 4)(x^2 + 7)') - ((x - 4)'(x^2 + 7)))/((x - 4)^2)
f'(x) = (((x - 4)(2x)) - (1(x^2 + 7)))/((x - 4)^2)
f'(x) = (2x^2 - 8x - x^2 - 7)/((x - 4)^2)
f'(x) = (x^2 - 8x - 7)/((x - 4)^2)

ANS : (x^2 - 8x - 7)/((x - 4)^2) or (x^2 - 8x - 7)/(x^2 - 8x + 16)

Answer 5

ley y = (X^2+7)/(X-4)
SO Y' =( (x-4)d/dx(x^2+7) - (x^2+7)d/dx(x-4) )/(X-4)^2
(USING FORMULA
d/dx(u/v) = (v.du/dx-u.dv/dv) /v^2
so y' = ((x-4)2x - (x^2+7)1)/(x-4)^2
=> y' = (2x^2 - 8x - x^2 - 7)/(x-4)^2
=> y' = (x^2-8x-7)/(x-4)^2
thats answer

Answer 6

just use u/v rule