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Explaining derivative graphically

daigo

Member
Jun 27, 2012
60
Here is f(x) = x^2:



And the derivative of it (2x):



So each point on the slope of the derivative is supposed to represent the slope of the line tangent at a certain point on the original function.

Say I choose an x-value on the derivative 1, so the point on the line would be (1,2).

Where on the original function would this be the represented slope of? As I understand it, the x-value 1 corresponds with a slope of 2, the x-value 2 corresponds with a slope of 4, etc. but how do I find the point on the original function where these are the slopes of?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Here is f(x) = x^2:



And the derivative of it (2x):



So each point on the slope of the derivative is supposed to represent the slope of the line tangent at a certain point on the original function.

Say I choose an x-value on the derivative 1, so the point on the line would be (1,2).

Where on the original function would this be the represented slope of? As I understand it, the x-value 1 corresponds with a slope of 2, the x-value 2 corresponds with a slope of 4, etc. but how do I find the point on the original function where these are the slopes of?
You have \(f(x)=x^2\), and \(f'(x)=2x\), so when \(x=1\) the slope of the tangent is \(f'(1)=2\), so the tangent at \(x=1\) is a line of the form:
\[y=2x+c\]
where the value of \(c\) can be determined as we know that \(y=f(1)=1^2=1\) is the y-coordinate of the point on the curve where \(x=1\). Hence \((1,1)\) is on the tangent, using this we get:
\[1=2+c \Rightarrow\ c=-1\]
so the tangent to the curve at the point where \(x=1\) is \(y=2x-1\).

CB
 

daigo

Member
Jun 27, 2012
60
Good explanation CB, thanks. I believe I understand now.

At the point (x,f(x)) the slope is f'(x) evaluated and the y-coordinate of derivative function is the slope of the line tangent at the point (x,f(x)) in the original function, and the entire equation of the line (not just the slope) is represented by what you explained. Is how I am understanding it. Also you can use the point-slope formula to determine the line, right?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Also you can use the point-slope formula to determine the line, right?
You can use any formula you want for the straight line. I personally use exclusively the $y=mx+b$ equation.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
... Also you can use the point-slope formula to determine the line, right?
As Ackbach says, you can use whatever method you are most comfortable with or is more convenient, so yes.