Discovering the Derivate Definition for f(x) = x/(2x-1)

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The discussion focuses on finding the derivative of the function f(x) = x/(2x-1) using the limit definition of a derivative. The calculation involves applying the limit as h approaches 0 to the difference quotient, ultimately simplifying to arrive at f'(x) = -1/(2x-1)^2. Participants emphasize the importance of continuing discussions in the same thread for clarity and suggest learning LATEX formatting for better mathematical expression. The conversation highlights the correct application of derivative rules and encourages ongoing engagement with complex topics. Overall, the thread serves as a resource for understanding derivative calculations.
ladyrae
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Thanks for your help...

How about this one?

Using the definition of derivate find f ` (x)

f(x) = x/(2x-1)

f ` (x) = lim h->0 [(f(x+h)) – (f(x))]/h

lim h->0 ([((x+h) / (2x+2h-1))-(x/(2x-1))]/h) . [((2x+2h-1)(2x-1)) / ((2x+2h-1)(2x-1))]

lim h->0 [(2x^2)-x+(2xh)-h-(2x^2)-(2hx)+x]/(h(2x+2h-1)(2x-1))

lim h->0 -h/(h(2x+2h-1)(2x-1)) = -1/(2x-1)^2
 
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If f(x)=x/(2x-1), then it is correct.
 
typo

yes...thanks
 
just a suggestion, ladyrae:
if you have more questions on this, don't open any new thread, continue to use this instead.
In addition, try to learn LATEX formatting, it's not very difficult..
 
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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