5x+5h/x2+h2+2xh+9

Question

f(x) = 5x/x^2+9 . . find the different quotient of F; that is, find f(x+h)-f(x)/h, h cannot = 0. For the following function and simplify answer.

Function: f(x) = 10/x^2

Answer 1

f(x)=5x/(x²+9); f(x+h)=5(x+h)/((x+h)²+9)=5(x+h)/(x²+2hx+h²+9).

f(x+h)-f(x)=5(x+h)/(x²+2hx+h²+9)-5x/(x²+9)=

(5(x+h)(x²+9)-5x(x²+2hx+h²+9))/[(x²+9)(x²+2hx+h²+9)]=

(5x³+45x+5hx²+45h-5x³-10hx²-45x)/(x⁴+2hx³+x²h²+18x²+18hx+9h²+81)=

(45h-5hx²)/((x⁴+18x²+81)+2hx³+x²h²+9h²).

Divide by h:

(45-10x²)/((x²+9)²+2hx³+x²h²+9h²).

In the denominator we can let h→0, so we have the gradient (derivative):

(45-5x²)/(x²+9)², which is the limit as h→0.

If f(x)=10/x², then f(x+h)=10/(x+h)²,

so f(x+h)-f(x)=10(1/(x+h)²-1/x²)=10(x²-(x+h)²)/(x²(x+h)²)=10(-2xh-h²)/(x⁴+2x³h+x²h²).

Divide by h:

10(-2x-h)/(x⁴+2x³h+x²h²).

When h→0:

-20x/x⁴=-20/x³.