Differentiation from first principles- - cant do at all

  • Thread starter pat666
  • Start date
  • #1
pat666
709
0

Homework Statement


see attachment- have to do this because i can't figure out how to do the notation in this part sorry... I have no idea where to go with this and probably need quite a bit of help with it--- thanks.


Homework Equations





The Attempt at a Solution

 

Attachments

  • first prince.png
    first prince.png
    24.7 KB · Views: 393

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
43,021
970
The problem says "use the definition of derivative
[tex]\lim\frac{\delta y}{\delta x}[/tex]

Do you know what that means?
[tex]\frac{\delta y}{\delta x}= \frac{y(1+ \delta x)- y(1)}{\delta x}[/tex]

[itex]y(1)= 1^2- 1= 0[/itex] and [itex]y(1+ \delta x)= (1+ \delta x)^2- (1+ \delta x)[/itex].
 
  • #3
pat666
709
0
hey - sorry bud i don't get any of that? - the lecture that we had talked for about 15s on this and I really don't understand it. more help would be GREATLY appreciated.
 
  • #4
36,487
8,456
The derivative definition is usually presented using upper-case delta, [itex]\Delta[/itex] rather than lower-case delta, [itex]\delta[/itex] as you have.

It might be helpful to use function notation, letting f(x) = y = x2 - x. The derivative of f at 1 can be written this way:
[tex]\lim_{\Delta x \to 0} \frac{f(1 + \Delta x) - f(1)}{\Delta x}[/tex]

The fraction gives the slope of a secant line between (1, f(1)) and (1 + [itex]\Delta x[/itex], f(1 + [itex]\Delta x[/itex])). The numerator gives the vertical change (rise) and the denominator gives the horizontal change (run). As [itex]\Delta x[/itex] approaches zero, the slope of the secant line approaches the slope of the tangent line.

Substitute for f(1) and f(1 + [itex]\Delta x[/itex]) in the limit formula above, simplify, and then take the limit.

If you still don't understand, your text should have an explanation of this and some examples.
 
  • #5
36,487
8,456
BTW, you should post calculus problems (like this one) in the Calculus & Beyond section.
 

Suggested for: Differentiation from first principles- - cant do at all

Replies
5
Views
259
Replies
4
Views
517
Replies
12
Views
616
Replies
4
Views
594
Replies
2
Views
542
Replies
15
Views
804
Replies
22
Views
980
Replies
10
Views
444
Top