C1 Differentiation question [SOLVED]Watch

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Thread starter 4 years ago
#1
A curve has equation y = x3/2 + 48x-1/2

a) Find the derivative of y with respect to x

b) Find the equation of the tangent to this curve at the point where x=1

I think I could do b easily but I don't understand part a. 1
4 years ago
#2
(Original post by radprincess)
A curve has equation y = x3/2 + 48x-1/2

a) Find the derivative of y with respect to x

b) Find the equation of the tangent to this curve at the point where x=1

I think I could do b easily but I don't understand part a. Can you do Part B, then? You might find Part B's answer useful to doing Part A.
1
4 years ago
#3
Part a) dy/dx = (times by power, then lower the power).

Part b) If I'm not mistaken (i haven't done this in a while) you simply use: y-y1 = m(x-x1)

Where x1 is the x value given, y1 is the y value when x = 1 (sub x=1 into the original equation). m is the gradient, sub x=1 into your part a Answer.

I hope this helps.
0
4 years ago
#4
(Original post by radprincess)
A curve has equation y = x3/2 + 48x-1/2

a) Find the derivative of y with respect to x

b) Find the equation of the tangent to this curve at the point where x=1

I think I could do b easily but I don't understand part a. Hi,

I hope this helps, it should be correct. Check wolfram alpha or something to be sure.
0
4 years ago
#5
(Original post by radprincess)
A curve has equation y = x3/2 + 48x-1/2

a) Find the derivative of y with respect to x

b) Find the equation of the tangent to this curve at the point where x=1

I think I could do b easily but I don't understand part a. Yeah, the workings are correct - hope it helps!
0
4 years ago
#6
For Part A, it's asking you to find the first derivative of the equation (dy/dx). To do this, just multiply the coefficient of each term by the power that x is raised to, and then take away 1 from that power (for example, if y=2x2-3x2/3, then dy/dx=4x-2x-1/3).

Part B can't be done without Part A, as you need to find the equation of the line via (y-y1)=m(x-x1), where m is the gradient and (x1,y1) is a point on the line. As you already know x1 (it is given that x=1), you can find y1 by substituting x=1 into the original equation. To find the gradient, m, substitute x=1 into your answer for Part A, then sub all those values into
(y-y1)=m(x-x1) and simplify to find the equation of the line.
0
Thread starter 4 years ago
#7
Thank you so much everyone!
0
4 years ago
#8
(Original post by Willis_l96)
*
You should not post full solutions.
0
4 years ago
#9
(Original post by keromedic)
You should not post full solutions.
I would usually agree, but this is Core 1 lmao
0
4 years ago
#10
(Original post by Willis_l96)
I would usually agree, but this is Core 1 lmao
Which means the OP is even less experience and would probably benefit more from being guided to the answer.
0
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