# 3 different derivatives?

1. Sep 20, 2007

### fk378

1. The problem statement, all variables and given/known data
Find the tangent line and the normal line to the curve y=(1+2x)^2 at the point (1,9).

2. Relevant equations
f'(x)= [f(x+h) - f(x)] / h ->plug in x=1
f'(x)= [f(x+h) - f(x)] / h ->plug in (x+h)
Power Rule
Power Rule (after factoring out the polynomial)

3. The attempt at a solution
I tried to solve it 4 different ways using the above 4 different approaches:

Plugging in x=1 for the derivative formula yields 12
Plugging in (x+h) for the derivative formula yields 4+8x
Using the power rule for (1+2x)^2 = 2(1+2x) = 2+4x (simplified version of 4+8x)
Using the power rule after factoring out the polynomial is 1+4x+4x^2 = 4+8x

Regarding the first attempt, I thought I could plug in the x-coordinate to get the tangent equation. Since when is that wrong?

Last edited: Sep 20, 2007
2. Sep 20, 2007

### Dick

You have to use both the power rule and the chain rule for the derivative of (1+2x)^2. You get 2*(1+2x)*d(1+2x)/dx=4*(1+2x).

3. Sep 20, 2007

### fk378

Well 4*(1+2x) = 4+8x = 2+4x which is what I got.

4. Sep 20, 2007

### Dick

Since when is 4+8x=2+4x?

5. Sep 20, 2007

### D H

Staff Emeritus
fk378, I rearranged your OP a bit to make a point.

Dick confirmed this is correct.
You forgot the factor of 2 when you take the derivative of 1+2x. With this factor of 2 you get the same result as above.
Obviously this is not a global result; it is only valid at x=1. One question: What is 4+8x evaluated at x=1?

6. Sep 20, 2007

### fk378

4+8x simplified = 2+4x

7. Sep 20, 2007

### fk378

But isn't the question asking what the slope is at x=1? So why wouldn't plugging in work?

8. Sep 20, 2007

### Dick

Yes. And 2 simplified is 1. So 2=1.

9. Sep 20, 2007

### D H

Staff Emeritus
I ask again, what is 4+8x evaluated at x=1?

10. Sep 20, 2007

### fk378

Sorry D H,
yes it equals 1. But if it's asking for the equation of the tangent line, what do I regard 12 as?

Dick,
2 in the above equations is a common factor that can be factored out. 2=1 has no such common factor.

11. Sep 20, 2007

### D H

Staff Emeritus
How in the world is 4+8x evaluated at x=1 equal to 1? (Answer: 4+8x evaluated at x=1 is 12.) How in the world is 4+8x = 2+4x? (Answer: Its not.)

You appear have some very basic understanding issues here. I think you need some face-to-face time with your teacher.

12. Sep 20, 2007

### fk378

Oh I did mean 12. Sorry I was just thinking about the problem...
Is it not possible to simplify the equation? Does that just apply to derivatives?

13. Sep 20, 2007

### D H

Staff Emeritus
You can say 2*(2+4x) = 4+8x. You cannot say 2+4x=4+8x. They are different equations. You can't just throw that factor of two into the ozone layer.