Urgent: One Extremely Basic Derivative Function

1. Jan 15, 2009

jacksonpeeble

1. The problem statement, all variables and given/known data
Find an equation of the tangent line to the curve y=2x-x^2 at the point (2, -4).

2. Relevant equations
y=2x-x^2

3. The attempt at a solution
I do not have any idea how to do any derivative, other than the equation to do so. Please note that we have been given the answer key - I already know that the solution is y+4=-10(x-2).

This is for a Precalculus exam (but I've been informed that derivatives are calculus). I have been absent for a while, which is when we learned about derivatives, so any background information would be nice as well, but the test is tomorrow.

2. Jan 15, 2009

psykatic

The slop of the tangent can be found by differentiating the given equation, let it be $$\frac {\delta y}{\delta x}.$$

Now, the formula to find the equation is simply,
$$\frac{y-y_1}{x-x_1}=~\frac {\delta y}{\delta x}.$$

where, $$x_1$$ and $$y_1$$ are the given points!

Substitute, and you'll get the answer, rightaway!

3. Jan 15, 2009

NoMoreExams

That's only for equations of the form y = m*x + b i.e. linear for some form of linear. In general dy/dx is not that.

4. Jan 15, 2009

jacksonpeeble

Thanks, but could somebody explain basic derivatives in general, please?

5. Jan 15, 2009

NoMoreExams

6. Jan 15, 2009

jacksonpeeble

Now I am ;-). I'll post back if I'm still confused. It would be really great if (if anyone has the time) somebody could post a walk-through on an example problem, though.

7. Jan 15, 2009

NoMoreExams

8. Jan 15, 2009

jacksonpeeble

If anyone is still following this thread: after checking with other students in the class, I found that the teacher had corrected the problem on the sheet in class - f(x) was supposed to equal 2x-x^3, which makes a major difference.

NoMoreExams - I understand that you're trying to be helpful, and I assume that you must get people that don't bother searching elsewhere before posting, but I do. I searched Google, and had the page up for Tangent Lines on Wikipedia before you provided the Derivatives link. I had even run several advanced searches on Physics Forums to try to find very similar problems to mine.

So even though I may be the exception, I find it frustrating when people tell me to check in other places when I, personally, use posting on a forum only as a last resort (I don't believe in swamping forums with repeats of previous topics).

9. Jan 15, 2009

NoMoreExams

But doesn't the 2nd page on the google search I linked give you the necessary tutorial you need to solve your problem?

10. Jan 15, 2009

jgens

The tutorial in NoMoreExams' link does cover the process needed to solve your problem.

11. Jan 16, 2009

psykatic

Didnt you try this method? It'll give the answer straightaway,

$$y=2x - x^3$$
$$\frac{\delta y}{\delta x}=~ 2-3x^2$$

Now, this is the equation for the slope, substituting value of 'x' you'll get the slope of the line to be -10!
Then have this value in the equation mentioned above to get the equation of the tangent!

P.S:NoMoreExams please note this.

12. Jan 16, 2009

NoMoreExams

What am I noting? That evaluating the derivative at a point will give you the slope of the tangent line? I kind of knew that already. My point was that the derivative is not the same as (y - y1)/(x - x1) as you wrote it.

13. Jan 16, 2009

psykatic

$$(y-y_1)=\frac{\delta y}{\delta x} (x-x_1)$$

This is the exact representation of the equation of the required equation!

14. Jan 16, 2009

NoMoreExams

Once again, that's for a tangent line in which case it's true that the derivative of the tangent line will be its slope. That's what derivative, or one of the things, a derivative represents the slope of your curve at that point i.e. the slope of the tangent line to your curve at a particular point. My point wasn't to say you're flat out wrong, it was to clarify your point.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook