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.
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