Some calculus exercises doubts

[Taturana]

Homework Statement

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

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

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.
$f(x) = \frac{1}{x - a}, a \in \mathbb{R} - \left \{ -2; 4 \} \right; x = -2, x = 4$

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

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 a lot, 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:

$$f(x) = x(3x-5)$$

I need to use the chain rule here, right?

$$f(x) = x u$$
$$u = 3x -5$$
$$\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}$$
$$\frac{df}{du} = x$$
$$\frac{du}{dx} = 3$$
$$\frac{df}{dx} = 3x$$

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:

$$f(x) = x(3x-5)$$

I need to use the chain rule here, right?

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

$$f(x) = x u$$
$$u = 3x -5$$
$$\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}$$
$$\frac{df}{du} = x$$
$$\frac{du}{dx} = 3$$
$$\frac{df}{dx} = 3x$$

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.