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Quick differentiation question

Hi, I was wondering how to go about differentiating a term such as x22\frac{x^2}{2} ?
Thanks.
You use the indices rule to change the fraction then you differentiate it

Wait sorry ignore that
(edited 8 years ago)
Reply 2
Original post by Zenzation
Hi, I was wondering how to go about differentiating a term such as x22\frac{x^2}{2} ?
Thanks.


Differentiation is a linear operator so:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \left(af(x)\right) = a\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left(f(x)\right)\end{equation*}



And:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \left( f(x) + g(x)\right) = \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left( f(x)\right) + \frac{\mathrm{d}}{\mathrm{d}x} \left(g(x)\right)\end{equation*}



Secondly, we have the rule that:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \left(x^{n}\right) = nx^{n-1}\end{equation*}



Can you combine the above to differentiate your term?

Hint: here, you'll want to use the first rule with a=12a = \frac{1}{2} and the third rule with n=2n = 2.
(edited 8 years ago)
Reply 3
Original post by Zacken
Differentiation is a linear operator so:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \left(af(x)\right) = a\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left(f(x)\right)\end{equation*}



And:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \left( f(x) + g(x)\right) = \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left( f(x)\right) + \frac{\mathrm{d}}{\mathrm{d}x} \left(g(x)\right)\end{equation*}



Secondly, we have the rule that:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \left(x^{n}\right) = nx^{n-1}\end{equation*}



Can you combine the above to differentiate your term?

Hint: here, you'll want to use the first rule with a=12a = \frac{1}{2} and the third rule with n=2n = 2.


Could I just differentiate the top resulting in 2x2\frac{2x}{2} and then be left with xx ?
Reply 4
Original post by Zenzation
Could I just differentiate the top resulting in 2x2\frac{2x}{2} and then be left with xx ?


Yes, that's essentially what I said about the derivative being a linear operator.
Reply 5
Original post by Zacken
Yes, that's essentially what I said about the derivative being a linear operator.


Ah thank you for clearing that up, I was trying to treat the whole thing as one. don't know why that didn't spring to mind in the first place lol.
Reply 6
Original post by Zenzation
Ah thank you for clearing that up, I was trying to treat the whole thing as one. don't know why that didn't spring to mind in the first place lol.


Yep, no worries! Glad it's clicked now. :smile:

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