Ugh. Differentiation help.

Okay. So I have these 4 questions and have no idea why my answers are wrong, howver I will only show you the first one as it take forever to write this out & I have used the same principle on all of them:
find the equation of the tangents to the curves at given points:
2x^2 +9 @ x=2
for this I did dy/dx = 2*2 x^2-1
so m = 4x
then I went back and used x=2 into the raw form
y=2x^2+9
2*(2^2) + 9
y= 8+9
y=17
therefore
17 = 4x + c
x=2
so
17 = 4*2 +c
17 = 8 + c
c=9
equation of tangent = y = 4x+9

You have to use y-y1 = m(x-x1) to get your equation

You forgot to find the value of the differential at x = 2.

Okay, so could someone quote me exactly where I made my mistake. It's late and I've done 25 pages of notes on differentiation, so I'm having a hard time comprehending the maths terminology as I'm so worn out.
Thanks

I bold coded in my previous reply to make it clear. You didn't substitute x = 2 into the differential to get the gradient.

I bold coded in my previous reply to make it clear. You didn't substitute x = 2 into the differential to get the gradient.

17=4x+c
x=2
4x = 4*2
17 = 8 + c
17-8 = c
c=9
I dont understand what you mean. I thought that is putting the x value into the differentiated gradient

yeah?
so you're saying this 8 is our new gradient? or just value of x * gradient?
I would have thought it is the result of
y=4x+c
we know x =2
and thus y=8 +c

but you're saying its
y=8x+c
so infact
17=16+c ?
c=1
y=8x +1

The differential gives you the general equation for the gradient at any point on the line. Substituting a value of x gives the gradient at that point.

Yes, y = 8x + 1.

The differential gives you the general equation for the gradient at any point on the line. Substituting a value of x gives the gradient at that point.

Yes, y = 8x + 1.

Yeah, I've been doing this for 5 hours straight now so am getting groggy. It's just clicked.
The general dx/dy gives you the equation of any goven point, however the tangent isn't any given point, it is a singular value, in this case x=2 y=17
so we have to then specialise (if that's correct), the dx/dy differentiation to that specific point. I completely get it now!

Btw, last question, I have to find the equation of a tangent to 1/(x^3), was wondering if you could help?
y=1/x^3
x= -1

Rearrange to get $y =x^{-3}$, then follow roughly the same method. Tell me if you get stuck.

Just updated my last reply to you, take a look and see if that looks right to you
I'm glad I got it too! Thanks for that btw

Btw, last question, I have to find the equation of a tangent to 1/(x^3), was wondering if you could help?
y=1/x^3
x= -1
I understand that we need to do
y = 1/-1^3
and thus y = -1
so our coordinates are (-1,-1)
then we differentiate to get
dx/dy
so:
1 * x^-3
-3 * x-4
then we have to times put -1 (x) into the -3/ x^4
where we get -3/1 so -3
and thus
(y) -1 = -3 +c
so c= 2?

Yeah it looks right.

You are almost there
Your point is (-1,-1)
Your m is -3

So y=-3x+c

Put your point into that and you will see that c=-4