The derivative of a function..?

Question

We're supposed to solve fully. . . I understand the concept but I am not getting the correct answer..
for y=2x(squared)-7, where x=-2

i subed this into the equation
f(x+h)-f(x)
--------------
h
(when the limit is approaching 0)

My final answer was 4x-14 . . . what do i do with the x=-2????? and the correct answer says: -8x-15

no i have not learnt it that way yet...we're doing everything the long way then he said he was going to teach us an easier way....:

heres what i did...
2(x+h)squared-7-2x(squared)-7
-------------------------------------
h
2(xsquared+2xh+2hsquared)-7-2xsquared-7
divided by h
canceled out the 2xsquared to give
4xh+2hsquared-7-7
divided by h
cancelled out the h to give
4x+2h-14
subed in zero to give
4x-14

Answer 1

Have you learned in your class yet about taking derivatives this way:

f(x) = ax^n

so f'(x) {the derivative of f(x)} = n*ax^(n-1) ?

That's how I'd approach this problem. So you have:

f(x) = 2x^2 - 7

f'(x) = 4x

f'(-2) = 4(-2) = -8

This doesn't agree with what you said the correct answer was, so I'm wondering if you wrote down the problem correctly?

Sorry, wish I could help more!

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EDITED TO ADD: Thanks for the additional information, easternahill. It's been almost 30 yrs since I did derivatives the way your teacher is having you do them, so I'm not sure I can dredge up how to do them that way! As your teacher said, he will be showing you an easier way soon (the way I did the problem), and then once you learn that way you never go back to this "x+h" way again.

I'll look at my old calculus book to see if I can refresh my brain, but if I don't come back you will know I wasn't successful! :-)

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OK kid, here we go, but my final answer is still -8. But here's the derivative using the x-h method and limits:

f'(x) = lim (h-->0) [ ( f(x+h) - f(x) ) / h ]

= lim(h-->0) [ (2(x+h)^2 - 7 - (2x^2 - 7) / h ]

= lim(h-->0) [ (2(x+h)^2 - 2x^2) / h ]

= lim(h-->0) [ (2x^2 + 4xh + 2h^2 - 2x^2) / h ]

= lim(h-->0) [ (4xh + 2h^2) / h ]

= lim(h-->0) [ (4x + 2h)]

= 4x

Then f'(-2) = 4(-2) = -8

Sorry, kid, that's the best I can do!

PS -- you got -14 as part of your answer because you took -7-7 instead of -7-(-7)!

Answer 2

The person who answered this previously is absolutely correct in every part of the answer. The only other possibility I can think of is that when you say for y=2x(squared)-7, you mean (2x)^2 - 7 instead of 2x^2 - 7. That would change things slightly, giving a derivative of 8x instead of 4x, but it still isn't going to give a derivative of -8x - 15. Do realize that "the back of the book" isn't always right. There can be mistakes there, and this is likely one of those. (... and when a value of "x" is given in a derivative problem, you do just plug that number into the derivative to get a number rather than a function for the final answer.)