Question

f(x) = { 3 x + 8 } /{ 5 x + 5 },
find f'( x )
find f'(3)

Answer 1

I assume you are referring to derivatives?

\(\displaystyle \frac{d}{dx}(\frac{h(x)}{g(x)})=\frac{g(x)h^{\prime}(x)-h(x)g^{\prime}(x)}{g(x)^2}\)

Does this make sense?

Answer 2

For your problem you have,

\(\displaystyle f(x)=\frac{h(x)}{g(x)}=\frac{3x+8}{5x+5}\)

Answer 3

Yep. I get the wrong answer when I try doing the derivitive

Answer 4

Ok, lets break this up shall we?

\(\displaystyle h(x)=3x+8\)
\(\displaystyle g(x)=5x+5\)

\(h^{\prime}(x)=~?\)
\(g^{\prime}(x)=~?\)

Answer 5

What is the derivative of 3x+8?

Answer 6

3?

Answer 7

Yep, and how about 5x+5?

Answer 8

5

Answer 9

Yeppers!

So now we have,

\(h(x)=3x+8\\
g(x)=5x+5\\
~\\
h^{\prime}(x)=3\\
g^{\prime}(x)=5\)

Now we can input those into this formula,

\(\displaystyle \frac{d}{dx}(\frac{h(x)}{g(x)})=\frac{g(x)h^{\prime}(x)-h(x)g^{\prime}(x)}{g(x)^2}\)

What would this look like?

Answer 10

(5x+5)(3)-(3x+8)(5)/(5x+5)^2

Answer 11

so then when you cancel out the (5x+5) from the numerator and denominator, it should be (3)-(3x+8) (5)/(5x+5) correct?

Answer 12

Oh, i think i figured it out. The (5x+5) isn't allowed to cancel right?

Answer 13

\(\displaystyle \frac{(5x+5)(3)-(3x+8)(5)}{(5x+5)^2}\)

This is a technically correct answer. I am sure you could multiply stuff out, and FOIL and such and get stuff to cancel.

Answer 14

{ (5x+5)(3)-(3x+8)(5) } / (5x+5)^2
Use distributive property and simplify the expression within the curly braces { } above.

Answer 15

@kquach1 : Your earlier cancelling out the (5x + 5) is not correct. You cancel it from the first expression which is correct. But the second expression does not have a (5x + 5) in the numerator and so the denominator will be (5x + 5)^2 for the second term.

Instead just simplify the numerator in austinL's expression above and leave the denominator as (5x + 5)^2 or you can make the denominator 25(x + 1)^2.
That will be your f'(x). Substitute x = 3 to evaluate f'(3).