Here's the question you clicked on:
haterofmath
find the the rate of change dy/dx where x=x0
\[y=-17; x _{0}=3\]
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
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
/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