# Thread: Did something wrong in my calculations, but I don't know what.

1. ## Did something wrong in my calculations, but I don't know what.

$\displaystyle \frac{\frac{5}{6(x+h)+3}-\frac{5}{6x+3}}{h}$

Should end up in the form $\displaystyle \frac{A}{(Bx+Ch+3)(Dx+3)}$

I got my answer in that form and the values are correct for A, B, C, and D. But when I use it to answer the rest of the questions I'm getting wrong answers. I think I did something wrong in the beginning when getting it in that form.

Can someone go through the steps and give me your answer to check with mine. I need to see what I did wrong.

A = -30
B = 6
C = 6
D = 6

Can you display your final answer in the form: $\displaystyle \frac{A}{(Bx+Ch+3)(Dx+3)}$

2. It seems good to me...

I'll take the numerator first.

$\displaystyle \frac{5}{6(x+h) + 3} - \frac{5}{6x+3} = \frac{5}{6x+6h + 3} - \frac{5}{6x+3}$

$\displaystyle = \frac{5(6x+3) - 5(6x+6h+3)}{(6x+6h + 3)(6x+3)}$

$\displaystyle = \frac{(30x+15) - (30x+30h+15)}{(6x+6h + 3)(6x+3)}$

$\displaystyle = \frac{30x+15 - 30x-30h-15}{(6x+6h + 3)(6x+3)}$

$\displaystyle = \frac{-30h}{(6x+6h + 3)(6x+3)}$

Now, the new fraction:

$\displaystyle \frac{\frac{5}{6(x+h) + 3} - \frac{5}{6x+3}}{h} = \frac{\frac{-30h}{(6x+6h + 3)(6x+3)}}{h}$

$\displaystyle = \frac{-30h}{h(6x+6h + 3)(6x+3)}$

$\displaystyle = \frac{-30}{(6x+6h + 3)(6x+3)}$

3. That's exactly what I did, but how did you get rid of the h in the denominator? You got an error on one of your final steps.

4. I missed a '}' in my post. I corrected it.

Remember that:

$\displaystyle \frac{(\frac{a}{b})}{c} = \frac{a}{b} \times \frac{1}{c} = \frac{a}{bc}$

5. Okay, I remember now. But I have to then use it to find $\displaystyle \lim_{h \to 0}$. I keep getting that wrong.

The other parts like f'(2), and f'(3) I've gotten right. f'(1) apparently is wrong and I did it exactly the way I did the others.

6. Ok, now I'm not sure what you are talking about. Seems beyond what I've learned...

But there is an 'h' in the function. As h tends to zero, 6h will tend to zero, and the function tends to:

$\displaystyle = \frac{-30}{(6x+ 3)(6x+3)}$

7. f'(1) and such means the derivative of the function when x is 1, so plug 1 into the function wherever there is an x. Sorry if you haven't learned it yet and I confused you.

Thanks for the answers you really helped me out. I had the right idea, but I was unsure about how to enter it.

8. Ok, I'm familiar with derivatives. It's the limit that I'm not familiar with.

Uh... do you differentiate with the h? Because, if we already take it as h tends to zero, then,

$\displaystyle f(x) = \frac{-30}{(6x+ 3)(6x+3)} = -30(6x + 3)^{-2}$

$\displaystyle f'(x) = 60(6x+3)^{-3} . 6 = \frac{360}{(6x+3)^3}$

9. $\displaystyle \lim_{h \to 0}$ was $\displaystyle \frac{-30}{(6x+ 3)(6x+3)}$ just as you said.

You substitute 0 in $\displaystyle \frac{-30}{(6x+6h + 3)(6x+3)}$ for h which leaves you with $\displaystyle \frac{-30}{(6x+ 3)(6x+3)}$ which was the correct answer. I didn't need to go any further for this particular exercise.