# C1 Differentiation question [SOLVED] Watch

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A curve has equation y = 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.

^{3}^{/2 }+ 48x^{-1/2 }a) Find the derivative of y with respect to xb) 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.

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#2

(Original post by

A curve has equation y = 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.

**radprincess**)A curve has equation y = x

^{3}^{/2 }+ 48x^{-1/2 }a) Find the derivative of y with respect to xb) 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.

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#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.

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.

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#4

**radprincess**)

A curve has equation y = x

^{3}

^{/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.

I hope this helps, it should be correct. Check wolfram alpha or something to be sure.

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#5

**radprincess**)

A curve has equation y = x

^{3}

^{/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.

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#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=2x

Part B can't be done without Part A, as you need to find the equation of the line via (y-y1)=m(x-x

^{2}-3x^{2/3}, 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-x

_{1}), where m is the gradient and (x_{1},y1) is a point on the line. As you already know x_{1}(it is given that x=1), you can find y_{1}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-y_{1})=m(x-x_{1}) and simplify to find the equation of the line.
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#9

(Original post by

You should not post full solutions.

**keromedic**)You should not post full solutions.

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#10

(Original post by

I would usually agree, but this is Core 1 lmao

**Willis_l96**)I would usually agree, but this is Core 1 lmao

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