# Homework Help: Difference Quotient Isn't working.

1. Jul 27, 2010

1. The problem statement, all variables and given/known data
Find the derivative of f using the differance quotient and use the derivative of f to determine any points on the graph of f where the tangent line is horizontal.

$$f(x)=3x^3-9x$$

2. Relevant equations

3. The attempt at a solution

$$\lim_{\Delta x\rightarrow0}=\frac{3(x+\Delta x)^3-9(x+\Delta x)-3x^3+9x}{\Delta x}$$

$$\lim_{\Delta x\rightarrow0}=\frac{3(x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3)-9x-9\Delta x-3x^3+9x}{\Delta x}$$

$$\lim_{\Delta x\rightarrow0}=\frac{3x^3+9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9x-9\Delta x-3x^3+9x}{\Delta x}$$

The '3x3's and '9x's cancel out and you are left with

$$\lim_{\Delta x\rightarrow0}=\frac{9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9\Delta x}{\Delta x}$$

Then if you take the limit as delta x approaches 0 you will get

$$9x^2+9x-6$$ I know the answer is $$9x^2-9$$ via the rules of differentiation.

What am I doing wrong with this example?

2. Jul 27, 2010

### Gib Z

Check the bolded step again.

3. Jul 27, 2010

### Dickfore

You can cancel factors of $\Delta x$ from each term in the numerator with the one in the denominator. What are you left with after this cancellation? What is the limit of this expression when $\Delta x \rightarrow 0$?

4. Jul 27, 2010

$$\lim_{\Delta x\rightarrow0}=\frac{9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9\Delta x}{\Delta x}$$

Ok, so you would get this after taking the limit.

$$=9x^2+9x\Delta x+3\Delta x^2-9$$

what happens to the delta x terms though? Is delta x just such a small number $$\epsilon$$ that you can say they are practically 0 and anything multiplied by 0 is 0?

5. Jul 27, 2010

### Dickfore

No. Please review limits before you go on to derivatives.

6. Jul 27, 2010

Nevermind you would simply factor out a delta x out of the numerator then you have 2 'delta x's that cancel out. Then you take the limit and you get 9x2-9. Otherwise its in indiscriminate form.

7. Jul 28, 2010

### HallsofIvy

I think you mean "indeterminant" form.

8. Jul 28, 2010

### Dickfore

I think you meant "indeterminate" form.

9. Jul 28, 2010

### Staff: Mentor

HallsOfIvy was being indiscriminate.

10. Jul 28, 2010

### Mentallic

Hahaha gold :rofl: