How Do You Differentiate x^2y and xy^2?

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x^2y

the y is biglike the x its not small

and how do you differentiate this
xy^2

for the first one i used the product rule and thesecond one and i get a different answer then what i was suppost to get
 
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for the first one i didthis

x^2 d/dx y dy/dx + y d/dx x^2

x^2 y' + y+ 2x

so how come I am not getting 2x 2y
 
:confused: You need an equation, not an expression, i.e. x^{2}y = ?. Also, are you familiar with implicit differentiation?
 
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ill give u the equation x^3+x^2y+4y^2=6
 
i understand it now, the answer of this was factored out or cleaned up a bit, so i didn't notice there was an x outside of the paranthesis of 2y
 
x^3 + x^2y + 4y^2 = 6

Okay, but do you know what implicit differentiation is? For example,

x + y = 1 [/itex]<br /> <br /> \frac{dx}{dx} + \frac{dy}{dx} = \frac{d1}{dx}<br /> <br /> 1 + \frac{dy}{dx} = 0<br /> <br /> Also for a product, the product rule applies, e.g.<br /> <br /> xy = 1<br /> <br /> x*\frac{dy}{dx} + \frac{dx}{dx}*y = 0
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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