# is there a way to simplify $x^{2} ( v' (x^{n})' )$?

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?

• derivatives are with respect to x by the way, and v is a function of x – Grant Dec 9 '14 at 4:58
• 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. – Ahaan S. Rungta Dec 9 '14 at 5:00
• 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? – Grant Dec 9 '14 at 5:01
• So you want to simplify $x^2 \cdot v'(x) \cdot \left( x^{n} \right)'$? – Ahaan S. Rungta Dec 9 '14 at 5:03
• yes exactly thats what i need – Grant Dec 9 '14 at 5:05

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 $v(x)$.