help please number 2

Question

marcelie · Answer

attachment

photon336 · Answer

@marcelie chose your u and your v first then differentiate both

mertsj · Answer

Rewrite it:
$$y=3x^2(4-2x)^{\frac{1}{2}}$$

marcelie · Answer

okay so it can be written like this 

u =  3x^2 
u'= 6x


v =  (4-2x)^1/2 
v'= 1/2(4-2x)^-1/5

marcelie · Answer

come back D: @Photon336

mertsj · Answer

$$v'=\frac{1}{2}(4-2x)^{-\frac{1}{2}}(-2)$$

photon336 · Answer

plug it into the formula for product rule $$v*(\frac{ du }{ dx })+u*(\frac{ dv }{ dx })$$

marcelie · Answer

marcelie · Answer

this is what i did

photon336 · Answer

$$6x*\sqrt{4-2x}-\frac{ 3x^{2} }{ \sqrt{4-2x} }$$

photon336 · Answer

yeah I saw what you did were you trying to simplify your answer further?

marcelie · Answer

yes i just got stuck on the top portion

photon336 · Answer

I'm trying to follow what you did in your notes. I'm getting something like this 

$$\sqrt{4-2x}*6x*\sqrt{4-2x}-\frac{ 3x^{2} }{ \sqrt{4-2x} }*\sqrt{4-2x}$$

$$\frac{ 6x(4-2x)-3x^{2} }{ \sqrt{4-2x} }$$

$$\frac{ 24x-12x^{2}-3x^{2} }{ \sqrt{4-2x} } = \frac{ 24x-15x^{2} }{ \sqrt{4-2x} }$$

marcelie · Answer

oh okay hmm  question if we are given something like this 

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