Some calculus exercises doubts

[Taturana]

Homework Statement

Problem 1
Find the equation of the tangent line on the following points of the curve.
[itex]f(x) = \frac{1}{x - a}, a \in \mathbb{R} - \left \{ -2; 4 \} \right; x = -2, x = 4[/itex]

Problem 2
Find the equation of the line normal to the tangent line on the following points of the curve.
[itex]f(x) = x^{2} - 1; x = 0[/itex]

Homework Equations

The Attempt at a Solution

Problem 1
[PLAIN]http://img21.imageshack.us/img21/4021/giflatex2.gif

As you can see my solution is completely wrong ;(
Problem 2
[PLAIN]http://img46.imageshack.us/img46/4265/giflatex3.gif

It will give a division by zero, what's going on here?

 

[Mark44]

Homework Statement

Problem 1
Find the equation of the tangent line on the following points of the curve.
[itex]f(x) = \frac{1}{x - a}, a \in \mathbb{R} - \left \{ -2; 4 \} \right; x = -2, x = 4[/itex]

Problem 2
Find the equation of the line normal to the tangent line on the following points of the curve.
[itex]f(x) = x^{2} - 1; x = 0[/itex]

Homework Equations

The Attempt at a Solution

Problem 1
[PLAIN]http://img21.imageshack.us/img21/4021/giflatex2.gif

As you can see my solution is completely wrong ;(



Problem 2
[PLAIN]http://img46.imageshack.us/img46/4265/giflatex3.gif

It will give a division by zero, what's going on here?

Evaluate your function and its derivative, and then use the values in your formulas. For #2, since f'(x) = 2x, then f'(0) = 0. This says that the tangent line is horizontal when x = 0 (i.e., at (0, 0). What does this say about the normal to the curve at this point?

 

[vela]

In problem 1, you simply forgot to substitute x=-2 in when you wrote in the expression for f'(-2) in the equation of the line.

In problem 2, you found the slope of the tangent to be given by f'(x)=2x. At x=0, you get f'(x)=0, which means the tangent line is horizontal. The normal, therefore, will be vertical. Vertical lines do not have a finite slope, so you can't use the point-slope formula for a line.

 

[Mark44]

For #1, evaluate f(-2) and f'(-2) before trying to find the equation of the tangent line. My answer agrees with the book's answer that you posted.

 

[Taturana]

Evaluate your function and its derivative, and then use the values in your formulas. For #2, since f'(x) = 2x, then f'(0) = 0. This says that the tangent line is horizontal when x = 0 (i.e., at (0, 0). What does this say about the normal to the curve at this point?

In problem 1, you simply forgot to substitute x=-2 in when you wrote in the expression for f'(-2) in the equation of the line.

In problem 2, you found the slope of the tangent to be given by f'(x)=2x. At x=0, you get f'(x)=0, which means the tangent line is horizontal. The normal, therefore, will be vertical. Vertical lines do not have a finite slope, so you can't use the point-slope formula for a line.

For #1, evaluate f(-2) and f'(-2) before trying to find the equation of the tangent line. My answer agrees with the book's answer that you posted.

Thank you for the answers, that helped me alot, now it works... =D

Just another question: how do I know when do I have to use the chain rule? There's any easy way of visualize it or is it just practice?

 

[Mark44]

Use the chain rule whenever you are differentiating a composite function. The chain rule actually comes into play in your first problem, but because the derivative of x - a is 1, the contribution from the chain rule isn't noticeable.

If you had f(x) = 1/(2x - a) = (2x - a)^-1, then f'(x) = -1(2x - a)^-2 * 2 = -2/(2x - a)². The 2 that appears in the numerator of the last expression comes from the use of the chain rule, and is the derivative of 2x - a.

 

[Taturana]

Use the chain rule whenever you are differentiating a composite function. The chain rule actually comes into play in your first problem, but because the derivative of x - a is 1, the contribution from the chain rule isn't noticeable.

If you had f(x) = 1/(2x - a) = (2x - a)^-1, then f'(x) = -1(2x - a)^-2 * 2 = -2/(2x - a)². The 2 that appears in the numerator of the last expression comes from the use of the chain rule, and is the derivative of 2x - a.

Thank you but I have another doubt (I think its not necessary to open another thread).

suppose:

[tex]f(x) = x(3x-5)[/tex]

I need to use the chain rule here, right?

[tex]f(x) = x u[/tex]
[tex]u = 3x -5[/tex]
[tex]\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}[/tex]
[tex]\frac{df}{du} = x[/tex]
[tex]\frac{du}{dx} = 3[/tex]
[tex]\frac{df}{dx} = 3x[/tex]

I know that is not right, but why?

 

[Mark44]

Thank you but I have another doubt (I think its not necessary to open another thread).

You have another question, not a doubt. You have a doubt about something when you think you know what it is, but you aren't sure.

And yes, you should start a new thread when you have a new problem.

suppose:

[tex]f(x) = x(3x-5)[/tex]

I need to use the chain rule here, right?

No. This is a product, not a composition, so you should use the product rule.

[tex]f(x) = x u[/tex]
[tex]u = 3x -5[/tex]
[tex]\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}[/tex]
[tex]\frac{df}{du} = x[/tex]
[tex]\frac{du}{dx} = 3[/tex]
[tex]\frac{df}{dx} = 3x[/tex]

I know that is not right, but why?

Here it is using the product rule.
f'(x) = x * d/dx(3x - 5) + (3x - 5) * d/dx(x)
= x * 3 + (3x - 5) * 1
= 3x + 3x -5 = 6x - 5.

As a check, f(x) = 3x² - 5x, so f'(x) = 6x - 5.