Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

Will fan if helped! Evaluate the derivative of the given function at the given point y=x^2 - x @ (3,6)

OpenStudy (anonymous):

True

OpenStudy (anonymous):

Easy, use the formula: (a+h)-F(a) -------- h

OpenStudy (anonymous):

And how would I incorporate it?

OpenStudy (anonymous):

I really appreciate the help

OpenStudy (anonymous):

plug in 6 in the a's

OpenStudy (anonymous):

You are correct pickle

OpenStudy (ranga):

Have you been taught derivative yet? If yes, then find dy/dx and evaluate it at x = 3.

OpenStudy (anonymous):

i mean plug it in the function.

OpenStudy (anonymous):

so y = 3x - 1?

OpenStudy (anonymous):

i mean 2x-1

OpenStudy (ranga):

y = x^2 - x dy/dx = 2x - 1 dy/dx at x = 3 is: 2 * 3 - 1 = 6 - 1 = 5

OpenStudy (anonymous):

(a+h)^2-(a+h)- a

OpenStudy (anonymous):

is that it ranga? what about ((a+h) - f(a))/h ?

OpenStudy (anonymous):

that equation confuses me

OpenStudy (anonymous):

Ok, are you confused about what you post, or if you are correct?

OpenStudy (anonymous):

just plug in (a+h) in the x's and then do (minus a) at the end of that equation because there's is - f(a) at the end of the formula

OpenStudy (ranga):

If you have not been taught derivatives yet, then you will have to do it the other way and evaluate the limit. But if you have been taught derivatives this is the easier way.

OpenStudy (anonymous):

ahh I see, thanks

OpenStudy (ranga):

You are welcome.

OpenStudy (anonymous):

tell me what you found

OpenStudy (anonymous):

y= 3x -1 x = 3 y= 3(2) - 1 y = 5

OpenStudy (anonymous):

Good job :)

OpenStudy (anonymous):

ok, it seemed a lot mroe complicated then it was since the equations seemed kind of daunting.

OpenStudy (anonymous):

but thanks

OpenStudy (anonymous):

The secret is to follow the formulas

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!