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# How to know when to differentiate/integrate watch

1. Basically what the heading says.

For example: Find the equation of the curve with the given derivative of y with respect to x that passes through the given point:

dy/dx = 3x2 + 2x point (2,10)

You find the equation via differentiation couldn't you, because m = 3(22)+2(2) =16

then use y-y1 =m (x-x1)

HOWEVER, I checked the answer and it does it by integrating, why? And how can I know to do this in the exam (basically what is the give away?)

*SIDE NOTE* Oh, wait, is it because differentiation is used for finding the equation of a line,not a curve?!
2. (Original post by rm_27)
Basically what the heading says.

For example: Find the equation of the curve with the given derivative of y with respect to x that passes through the given point:

dy/dx = 3x2 + 2x point (2,10)

You find the equation via differentiation couldn't you, because m = 3(22)+2(2) =16

then use y-y1 =m (x-x1)

HOWEVER, I checked the answer and it does it by integrating, why? And how can I know to do this in the exam (basically what is the give away?)

*SIDE NOTE* Oh, wait, is it because differentiation is used for finding the equation of a line,not a curve?!
You've already been given the derivative of the curve so all that's needed involving differentiation is to evaluate it by substituting the value of x into it (2) to get the gradient of it- 16 (m). Then like you said, i would use the formula you mentioned (y-y^1=...). Perhaps you can go about it using integration, but as long as you get the same equation, using differentiation is fine.
3. (Original post by rm_27)
...
When you calculate , that is gradient of curve at that given point.
Method you employed is for finding equation of tangent at that same point.

They want the equation of curve itself.
The derivative given in question comes from differentiating the curve so integrate to reverse.
4. (Original post by rm_27)
Basically what the heading says.

For example: Find the equation of the curve with the given derivative of y with respect to x that passes through the given point:

dy/dx = 3x2 + 2x point (2,10)

You find the equation via differentiation couldn't you, because m = 3(22)+2(2) =16

then use y-y1 =m (x-x1)

HOWEVER, I checked the answer and it does it by integrating, why? And how can I know to do this in the exam (basically what is the give away?)

*SIDE NOTE* Oh, wait, is it because differentiation is used for finding the equation of a line,not a curve?!
Your 'SIDE NOTE' is wrong. A curve,line,circleare all functions so differentiation can be used for them.

Usually you have a equation (eg y=2x+3) and to find the gradient, you do dy/dy=2. Likewise, if you are given dy/dx as 2, to find the original line (to go back to y=2x+3), you use integration. Integration is the 'anti' of differentiation. If you integrate my example, you get y=2x+c. You use the coordinates they give you in the question such as when x=0,y=3 to find the c as 3. So in general if you know dy/dx, integrate to find the function and sub in the coordinates for c if told. Just do more practice questions and hopefully you get the concept

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Updated: December 27, 2014
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