find dy/dx of the following functions :

y=(3+x/3-x)^4 anyone?

Question

find dy/dx of the following functions :

y=(3+x/3-x)^4 anyone?

mertsj · Answer

$$y=(\frac{3+x}{3-x})^4$$

mertsj · Answer

First use the power rule.  Then use the quotient rule on the fractional part.

cucacula · Answer

need to find U and V first? @Mertsj

e.mccormick · Answer

Well, there would be a chain rule for the overall.  The outer power is one part, and everything inside the paren is the other.

cucacula · Answer

so the U = 3 + x and V = 4 - x ? @e.mccormick

e.mccormick · Answer

Well, the part inside the paren can be done using the quotent rule, but the chain rule must be applied to the overall.  They use u and v all over calc, so what is u and what is v depends on which rule you are talking about.

anonymous · Answer

$$y= (\frac{ 3 + x }{ 3 - x })^{4}$$ can be written as $$y= \frac{ (3 + x)^{4} }{ (3 - x)^{4} }$$

anonymous · Answer

the quotient rule (differentiation) says that when you have $$y = \frac{ u }{ v }$$ and you differentiate it,
$$\frac{ dy }{ dx } = \frac{ (v \times \frac{ du }{ dx })-(u \times \frac{ dv }{ dx }) }{ v ^{2} }$$

cucacula · Answer

but need to find U and V again? @kausarsalley

anonymous · Answer

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