Differeniate
V^3-2v√v/v
I used the quotient rule
(v)(v^3-2v√v)' -(v^3-2v√v)(v)' /(v)
The answer is 2v-1-1
I got this far but then I don't know what to do further
2v^3-2v^1/2+2v√v/(v)^2
looks like a continuation of ...
http://www.jiskha.com/display.cgi?id=1455252502
now you changed it to
V^3-2v√v/v
which clearly reduces to
V^3-2√v
and the derivative is even easier
f(v) = v^3 - 2v^(1/2)
f ' (v) = 3v^2 - v^(-1/2)
or
3v^2 - 1/√v
unless she really meant
(v^3-2v√v)/v
which is just
v^2 - 2√v
and integrates to
v^3/3 - 4v√v/3
+c
To differentiate the expression V^3 - 2V√V / V using the quotient rule, you're on the right track. As you correctly stated, the quotient rule states:
(f/g)' = (f'g - fg') / (g^2)
Let's apply this rule step by step to differentiate the given expression:
Step 1:
Identify f and g in the expression:
f = V^3 - 2V√V
g = V
Step 2:
Find the derivatives of f and g:
f' = d/dV(V^3 - 2V√V)
= 3V^2 - 2(1/2)(V^(-1/2))(√V)
= 3V^2 - V^(-1/2)√V
= 3V^2 - √V
g' = d/dV(V)
= 1 (since the derivative of V with respect to V is just 1)
Step 3:
Apply the quotient rule:
(f/g)' = (f'g - fg') / (g^2)
= ((3V^2 - √V) * V - (V^3 - 2V√V) * 1) / V^2
Simplifying this expression further:
= (3V^3 - V - V^3 + 2V^3/2) / V^2
= (2V^3 + 3V^3 - V^3 - V) / V^2
= (4V^3 - V) / V^2
= 4V - (V/V^2)
= 4V - 1/V
Therefore, the derivative of the expression V^3 - 2V√V / V is given by 4V - 1/V.