Forh≠ 0, which of the following is equivalent to

$\displaystyle

(A(x+h)^2-Ax^2)/h

$?

i have 5 answers...

a) 2Ax + Ah

b) Ax + Ah

c) 2Ax + Ah^2

d) Ah

e) 2Ax

How would I go about solving this problem?

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- Aug 31st 2009, 04:02 PMJAMESveederAlgebra problem...
For

*h*≠ 0, which of the following is equivalent to

$\displaystyle

(A(x+h)^2-Ax^2)/h

$?

i have 5 answers...

a) 2Ax + Ah

b) Ax + Ah

c) 2Ax + Ah^2

d) Ah

e) 2Ax

How would I go about solving this problem? - Aug 31st 2009, 04:39 PMartvandalay11
expand the numerator

$\displaystyle (x+h)^2=x^2+2h+h^2$ now multiply that by A, combine like terms, factor to cancel an h, and you'll have your answer - Aug 31st 2009, 04:44 PMpomp
$\displaystyle \frac{(A(x+h)^2 - Ax^2)}{h} = \frac{A(x^2 + 2xh + h^2) - Ax^2}{h} $

$\displaystyle = \frac{A(x^2 + 2xh + h^2 - x^2)}{h} $

You should be able to take it from here.

If you didn't already know, it seems this question was asked with the aim of introducing you to calculating derivatives from first principles, in this case the derivative of Ax^2. To find the derivative of the function you take the limit as h tends to zero:

$\displaystyle \frac{d}{dx} Ax^2 = \lim_{h \to 0} \frac{(A(x+h)^2 - Ax^2)}{h} = \lim_{h \to 0} (2Ax + Ah) = 2Ax$

as you would expect.

Hope this helped,

pomp.