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so I have what is probably an algebra question, if I have $x^{2} ( v' (x^{n})' )$ where the ' denotes a derivative, is there a way to simplify this expression?

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  • $\begingroup$ derivatives are with respect to x by the way, and v is a function of x $\endgroup$ – Grant Dec 9 '14 at 4:58
  • $\begingroup$ I assume you want $$ x^2 \cdot \frac {\mathrm{d}v}{\mathrm{d}x} \left( x^n \right)? $$ If this is the case, I'll provide an answer. $\endgroup$ – Ahaan S. Rungta Dec 9 '14 at 5:00
  • $\begingroup$ actually the x^n is also differentiated!I ask beacuse I know x^2 * x^n is x^n+2 but is there something similar I can do in the above case? $\endgroup$ – Grant Dec 9 '14 at 5:01
  • $\begingroup$ So you want to simplify $ x^2 \cdot v'(x) \cdot \left( x^{n} \right)' $? $\endgroup$ – Ahaan S. Rungta Dec 9 '14 at 5:03
  • $\begingroup$ yes exactly thats what i need $\endgroup$ – Grant Dec 9 '14 at 5:05
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Note that $$ \begin {align*} x^2 \cdot v'(x) \cdot \left( x^n \right)' &= x^2 \cdot v'(x) \cdot nx^{n-1} \\&= nx^{n+1} \cdot v'(x), \end {align*} $$ which is as far as you can get if you don't know $v(x)$.

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