How does one find the gradient of a curve at a specified point?

For example, my question is





find the gradient of the graph of y=4-5x^2-x^3 at the point (-2,-8)





i tried doing this by differentiating the graph formula and the substituting x... and eventually recieved the gradient -22 =s





am i on the right track atall?How does one find the gradient of a curve at a specified point?
gradient is the differential of the curve.





dy/dx=-10x-3x^2





at x=-2 dy/dx=(-10*-2)-(3*(-2)^2)=8How does one find the gradient of a curve at a specified point?
s = dy/dx = -10x - 3x^2


s(-2) = -10(-2) -3(-2)^2 = 20 - 12 = 8





(can't pinpoint your error without seeing your work)
y=4-5x^2-x^3


y'=-10x-3x^2


y'(-2)=-10(-2)-3(-2)^2=20-12=8





I don't know how you got -22. Check your arithmetic, but you seem to be on the right track.
You probably did the right method, but messed up the calculations. If by gradient you mean slope, then the first derivative will give you the answer:





y'= -10*x-3*x^2





Replacing x with the point you were given:





y'= -10(-2)-3(.2)^2 = 20-12 = 8





which is the slope at x=-2.
Let y = f(x)





Then let the gradient of f be defined as f'.....ok?





Then f(x) = 4 - 5x^2 - x^3





==%26gt; f'(x) = -10x - 3x^2





Ya with me? If you're at this level you better be saying ';yeah';....lol.





Then for x = (-8), you'd pretty much better get y = f(-8) = (-2)





Let's see if it does:





f(-2) = -10(-2) - 3(-2)^2





==%26gt; 20 - 3*4


==%26gt; 20 - 12 = -8





So, it does work out. This means the specified point (-2, -8) does lie on the gradient curve.





This is how you do these, and the ';gradient'; just means that you are taking the derivative of a function, and when you are looking at one specific point, you are evaluating that function at that point.





This is extremely basic stuff!!! In fact if this question has stumped you, either one of two things is going on. You've not doing your own work, or you are in over your head my friend.





??
z= y - 4+5x^2 + x^3


the gradient of f is %26lt;10x+3X^2, 1%26gt;


u = -2i + -8j


Dsub u (if u wanna fin the directional derivative of f at that point) = gradient of f * u

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