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I have been attempting to wrap my head around this problem for a couple days now.

I've attempted numerous different iterations to try and find how the answer is derived, but I just don't see the connection.

Any help on this matter would be greatly appreciated. Thanks.

Question: $$3x^5-5y^3 = 5x^2+3y^5\text{. Find }\frac{\mathrm{d}y}{\mathrm{d}x}$$

Book's answer: $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x(3x^3-2)}{3y^2(y^2+1)}$$

If you could show the steps to get said answer, that would be most helpful. Thanks.

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We differentiate the entire expression with respect to $x$

$$3x^5-5y^3=5x^2+3y^5$$

becomes

$$15x^4-15y^2\frac{\mathrm{d}y}{\mathrm{d}x}=10x+15y^4\frac{\mathrm{d}y}{\mathrm{d}x}$$

Rearranging to solve for $\frac{\mathrm{d}y}{\mathrm{d}x}$ and factoring out we get

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{15x^4-10x}{15y^4+15y^2}=\frac{5x(3x^3-2)}{15y^2(y^2+1)}=\frac{x(3x^3-2)}{3y^2(y^2+1)}$$

Let me know if there was any particular step you didn't understand

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  • $\begingroup$ That makes perfect sense! Separating the X's and Y's was the step we weren't taught, and that was the key! Thanks! $\endgroup$ – Sean May 17 '15 at 18:19
  • $\begingroup$ @Sean No worries, click the correct answer tick if it has solved your problem. $\endgroup$ – Sam Houston May 18 '15 at 14:03

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