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# Differentiation/Tangent watch

1. Given that y=3xSquared + 4
Find Dy/Dx.

I got that as 6x.

Then find the gradient of the curve when x=3.

I got this as x=18.

Then find the tangent to the curve at this point.

I need help here. I understand you use y=mx+c and therefore i have y=18x+c ?

Thanks
Xphoenix
2. (Original post by Xphoenix)
I need help here. I understand you use y=mx+c and therefore i have y=18x+c ?
Yes. You need to find a point on the tangent line, to allow you to calculate c. However.

1. the tangent line and the curve itself meet at a single point so this point is on both
2. you can find the coords of this point as you have a value for x, and an equation for the curve.
3. (Original post by Xphoenix)
Given that y=3xSquared + 4
Find Dy/Dx.

I got that as 6x.

Then find the gradient of the curve when x=3.

I got this as x=18.

Then find the tangent to the curve at this point.

I need help here. I understand you use y=mx+c and therefore i have y=18x+c ?

Thanks
Xphoenix
Typo perhaps but m=18 rather than x=18

Use the original equation to find y when x=3 then use that point in your y=18x+c
4. (Original post by Xphoenix)
Given that y=3xSquared + 4
Find Dy/Dx.

I got that as 6x.

Then find the gradient of the curve when x=3.

I got this as x=18.

Then find the tangent to the curve at this point.

I need help here. I understand you use y=mx+c and therefore i have y=18x+c ?

Thanks
Xphoenix
You need to the coordinates of the tangent also to put into y-y1=m (x-x1)

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