Difference Quotient Isn't working.

[themadhatter1]

Homework Statement

Find the derivative of f using the difference quotient and use the derivative of f to determine any points on the graph of f where the tangent line is horizontal.

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

Homework Equations

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?

 

[Gib Z]

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

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?

Check the bolded step again.

 

[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 $\Delta x \rightarrow 0$?

 

[themadhatter1]

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 $\Delta x \rightarrow 0$?

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

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?

 

[Dickfore]

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

 

[themadhatter1]

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.

 

[HallsofIvy]

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.

I think you mean "indeterminant" form.

 

[Dickfore]

I think you mean "indeterminant" form.

I think you meant "indeterminate" form.

 

[Mark44]

I think you mean "indeterminant" form.

I think you meant "indeterminate" form.

HallsOfIvy was being indiscriminate.

 

[Mentallic]

Hahaha gold