Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

find the the rate of change dy/dx where x=x0

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

\[y=-17; x _{0}=3\]
If the function is the constant function y = -17, then the derivative of it (dy/dx) is simply 0. The derivative of any constant is "0". Remember that "derivative" means "the change y with respect to x" in this case. If y = -7 at all points, then it never changes with respect to x. Therefore, the change is 0.
so what do you do with the 3

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

there has to be more to this question which you have not given us
if the equation for y is simply: y = -17 .. then daru is correct.
sorry wrong problem \[y=6-2x; x_{0}=3\]
first step, is to take the derivative of y, but the method of that step differs depending on how youre spose to go about it. there is a first principals method, and the derivative rules method. Which method have you been doing in your lessons?
derivative rules are much easier to work with then; what derivative rule would you say can apply to this question?
wouldn't the derivative be 2
almost, that negative sign (subtraction) needs to be recognized :) y= 6 - 2x y' = 6' - (2x)' = 0 - 2
now, since there is no place to plug in the value of x in the derivative; the value of the derivative at ANY point is: -2
oh ok.
/so how do you find that slope of tangent after taking the derivative?
the value of the derivative DEFINES the slope of the tangent at a given point.
in this case, 6 - 2x is a straight line whose sloe is -2 at all points
so in a problem such as f(x)=5x-3 the slope of the line that is tangent is 5?
the more complicated the curve, the less trivial the derivative becomes :) take: y=x^2 y' = 2x the slope at any given point of y=x^2 is defined as 2x the slope at x=3 would be 2(3) = 6 the slope at x=7 would be 2(7) = 14 the slope at x=-2 would be 2(-2) = -4
ok. thanks

Not the answer you are looking for?

Search for more explanations.

Ask your own question