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I need to find $y'$ from the equation $x^4y+\sqrt(xy)=5$ Wolffram and Mathway(pro) are giving me way different answers and I can not follow what they are doing and where I am going wrong.


My attempt:

$$x^4y+\sqrt{xy}=5$$

$$x^4y+(xy)^{1/2} = 5$$

$$x^4y'+y(4x^3)+\frac{1}{2}(xy)^\frac{-1}{2}(xy'+y)=0$$

$$y'(x^4+\frac{1}{2}(xy)^{1/2}(x+y))=-4x^3y$$

$$y' = -4x^3y / (x^4+1/2(xy)^{1/2}(x+y))$$

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    $\begingroup$ The mistake is in taking $y'$ outside the brackets in line 3-4. We have $x^4 y' + 4x^3 y + \frac{1}{2}(xy)^{-\frac{1}{2}}(xy' + y) = y'\left[x^4 + \frac{1}{2}(xy)^{-\frac{1}{2}}x\right] + 4x^3 y + \frac{1}{2}(xy)^{-\frac{1}{2}}y = 0$ $\endgroup$ – Winther Jun 3 '15 at 23:24
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You also can use implicit differentiation $$F=x^4y+\sqrt{xy}-5=0$$ $$F'_x=4 x^3 y+\frac{y}{2 \sqrt{x y}}$$ $$F'_y=x^4+\frac{x}{2 \sqrt{x y}}$$ So, after simplfications, $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{8 x^2 (x y)^{3/2}+y}{2 x^4 \sqrt{x y}+x}$$

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Use multiplication rule:

$$(x^4y)'=4x^3y+x^4y'$$

$$(\sqrt{xy})'=\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}$$

$$5'=0$$

We get

$$4x^3y+x^4y'+\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}=0$$

$$x^4y'+\frac{xy'}{2\sqrt{xy}}=-4x^3y-\frac{y}{2\sqrt{xy}}$$

$$y'\bigg(x^4+\frac{x}{2\sqrt{xy}}\bigg)=-4x^3y-\frac{y}{2\sqrt{xy}}$$

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  • $\begingroup$ Thank you for answering, but could you please go over how you found the derivative of sqrt(xy)? Maybe I am missing something because I thought I could put it all in parenthesis and to the power of 1/2 and then use the power rule for rational exponents? Which is pretty much the chain rule? $\endgroup$ – Jake Mager Jun 3 '15 at 23:41
  • $\begingroup$ $$((x)^{1/2})'=\frac12(x)^{-1/2}=\frac{1}{2\sqrt x}$$ Does this answer it? $\endgroup$ – grdgfgr Jun 4 '15 at 0:02
  • $\begingroup$ ohhhhh yes thank you $\endgroup$ – Jake Mager Jun 4 '15 at 1:12

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