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

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To find the derivative of the function \(4x - x^2\), the correct approach involves using the limit definition of the derivative. The initial attempt at applying the difference quotient was flawed due to algebraic errors, particularly in simplifying the numerator. The discussion highlights the importance of careful algebraic manipulation to avoid missing terms. Additionally, the Power Rule could be used for a more straightforward solution, although it seems the user may not be familiar with it yet. Overall, the conversation emphasizes the need for accuracy in calculations when determining derivatives.
kwikness
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Homework Statement


Find the Derivative Function of (4x - x^{2})


The Attempt at a Solution



using formula:

<br /> \frac{dy}{dx} = \frac{f(x + \Delta x) - f(x)}{\Delta x}<br />



<br /> \frac{4(x + \Delta x) - (-x^{2})}{\Delta x}<br />

<br /> \frac{4x + 4(\Delta x) + x^{2}}{\Delta x}<br />

Not sure where to go from here..
 
Last edited:
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Let d = \Deltax.

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
<br /> \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}<br />
as what you gave is just the difference quotient
<br /> \frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}<br />

Not sure how you got there in the first place. If I plug f(x) = 4x - x^2 into the formula you gave, I get
\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}<br />
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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