Derivative Check: Xe^{lnX^2}' = X^2e^{lnX^2}+2Xe^{lnX^2}

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\frac{d}{dx} Xe^{lnX^2} = e^{lnX^2}+Xe^{lnX^2} \frac{1}{X^2} 2x
=\frac{X^2 e^{lnX^2}+Xe^{lnX^2}2X}{X^2}

is that correct?
 
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True enough, but since e^{ln(x^{2})}=x^{2} there is an easier way to solve this..
 
oh wow...

3x^2

is that it?
 
Right..:wink:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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