Calculating Derivatives: How Many Different Ways Can You Do It?

[fk378]

Homework Statement

Find the tangent line and the normal line to the curve y=(1+2x)^2 at the point (1,9).

Homework 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)

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+8xRegarding the first attempt, I thought I could plug in the x-coordinate to get the tangent equation. Since when is that wrong?

 

[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).

 

[fk378]

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

 

[Dick]

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

 

[D H]

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

I tried to solve it 4 different ways using the above 4 different approaches:
Plugging in (x+h) for the derivative formula yields 4+8x
Using the power rule after factoring out the polynomial is 1+4x+4x^2 = 4+8x

Dick confirmed this is correct.

Using the power rule for (1+2x)^2 = 2(1+2x) = 2+4x

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.

Plugging in x=1 for the derivative formula yields 12

Obviously this is not a global result; it is only valid at x=1. One question: What is 4+8x evaluated at x=1?

 

[fk378]

4+8x simplified = 2+4x

 

[fk378]

Obviously this is not a global result; it is only valid at x=1. One question: What is 4+8x evaluated at x=1?

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

 

[Dick]

4+8x simplified = 2+4x

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

 

[D H]

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

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

 

[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.

 

[D H]

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.

 

[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?

 

[D H]

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.