How Do You Solve the Derivative of \(4x - x^2\)?

kwikness
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Homework Statement


Find the Derivative Function of (4x - x[tex]^{2}[/tex])


The Attempt at a Solution



using formula:

[tex] \frac{dy}{dx} = \frac{f(x + \Delta x) - f(x)}{\Delta x}[/tex]



[tex] \frac{4(x + \Delta x) - (-x^{2})}{\Delta x}[/tex]

[tex] \frac{4x + 4(\Delta x) + x^{2}}{\Delta x}[/tex]

Not sure where to go from here..
 
Last edited:
Let d = [itex]\Delta[/itex]x.

f(x+d) - f(x) = 4(x+d)-(x+d)^2 - 4x + x^2

which simplifies to (4 - d - 2x)d (you should derive this). The rest should be easy.
 
First of all, you mean
[tex] \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}[/tex]
as what you gave is just the difference quotient
[tex] \frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}[/tex]

Not sure how you got there in the first place. If I plug [itex]f(x) = 4x - x^2[/itex] into the formula you gave, I get
[tex]\frac{ [ 4(x + \Delta x) - (x + \Delta x)^2 ] - [4 x - x^2 ] }{ \Delta x }<br /> = \frac{ 4 x + 4 \Delta x - x^2 - 2 x \Delta x - (\Delta x)^2 - 4 x + x^2 }{ \Delta x}[/tex]
which has some terms you don't have. Now try again.
 
That's because you haven't finished the algebra! You have 4x and -4x in the numerator! You have -x2 and x2 in the numerator!
 
HallsofIvy said:
That's because you haven't finished the algebra! You have 4x and -4x in the numerator! You have -x2 and x2 in the numerator!

Exactly, and the -4x and -x2 just happen to be some of the terms kwikness is missing :smile:
But I'm leaving him some work.
 
Thanks, when I wrote it down I was missing a part of the equation. Gahhh! I always make stupid mistakes like that.
 
You could just have used the Power Rule, but I guess you haven't learned it yet.
 

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