Why Does My Calculated Tangent Line Not Touch the Curve at the Given Point?

[danne89]

Hi. I can't get what I'm doing wrong here. If somebody points that out, I'm really gratefull.

u = (6 + 2x^2)^3
d(u) = 3(6+2x^2)^2 d(6+2x^2)=3(6+2x^2)^2 d(6) + d(2x^2) <br /> = [3(6+2x^2)^2 * 0 + 4x]dx = 4xdx
Then I want to draw a tangent on the point, which acctually lies on the line(!), (1, 512) to check the derivative.
l(x)=f&#039;(a)(x-a)+b=4(x-1)+512=4x+508 Which don't intersect the curve at (1, 512)...

Please correct every misstake I've made. Yes, they could be many; I'm not so good on this stuff.

 

[matt grime]

You've forgotten that there is a bracket around 6+2x^3 in the second line of the maths.

You've misspelled mistake too. And who said irony is dead?

Anyway, du/dx is not 4x as should be obvious to you (u if you multiplied it out would have an x^6 term in it and hence du/dx must have an x^5 term in it.

 

[danne89]

Hmm. I'm sorry, but I don't get it. Can you please be a little more specific. This stuff is doing me crazy!

 

[matt grime]

Go through the second line of maths look how you write, in effect:

A(B+C) = AB + C.

when you go "across" the second equals sign in the line, ie from

(6+2x^2)^2 d(6+2x^2)=3(6+2x^2)^2 d(6) + d(2x^2)

they aren't equal. Your algebraic manipulation is wrong.

 

[danne89]

Ahh! Thanks! That should I've noticed...
d(y)=3[(x^2+5)^2][d(x^2+5)]=3[(x^2+5)^2][d(x^2)+d(5)] = 3[(x^2+5)^2]2x=6x(x^2+5)^2

 

[HallsofIvy]

Hi. I can't get what I'm doing wrong here. If somebody points that out, I'm really gratefull.

u = (6 + 2x^2)^3
d(u) = 3(6+2x^2)^2 d(6+2x^2)=3(6+2x^2)^2 d(6) + d(2x^2)

Should be
d(u)= 3(6+2x^2)^2 d(6+2x^2)= 3(6+2x^2)^2[d(6)+ d(2x^2)]
d(u)= 3(6+2x^2)^2[0+ 4xdx]= 12x(6+2x^2)^2dx.

Then I want to draw a tangent on the point, which acctually lies on the line(!), (1, 512) to check the derivative.
l(x)=f&#039;(a)(x-a)+b=4(x-1)+512=4x+508 Which don't intersect the curve at (1, 512)...

Please correct every misstake I've made. Yes, they could be many; I'm not so good on this stuff.

What you give: l(x)= 4x+ 508 certainly does intersect the curve at (1, 512): l(1)= 4+ 508= 512. It just isn't tangent to the curve because your f'(1) is incorrect.

When x= 1, du/dx= 12(1)(8)²= 768.
The tangent line should be y= 768(x-1)+ 512= 768x- 256.

That works nicely. (Aren't graphing calculators wonderful!)

 

[danne89]

What you give: l(x)= 4x+ 508 certainly does intersect the curve at (1, 512): l(1)= 4+ 508= 512. It just isn't tangent to the curve because your f'(1) is incorrect.

When x= 1, du/dx= 12(1)(8)²= 768.
The tangent line should be y= 768(x-1)+ 512= 768x- 256.

That works nicely. (Aren't graphing calculators wonderful!)

Mine doesn't give me that though.