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$y=6/(5-2x)$ I got $\frac{dy}{dx}=-6(5-2x)^{-2}$, but the answer is $12/(5-2x)^2$, how?

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    Don't forget chain rule. You pick up a factor of $2$ from the $2x$ in the denominator. – 2017-01-27
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    @CameronWilliams i replaced (5-2x) with u, getting 6*u^-1, then I derived this as -6u^-2, then replaced u with (5-2x) again, is that wrong? – 2017-01-27
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    Yes, you have to include a factor of $du/dx$ (which is $-2$) when you do this. Remember, $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$ -- known as the chain rule. – 2017-01-27
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    It's a simple mistake. Just multiply by the derivative of $5 - 2x$ when you're done. – 2017-01-27

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write your term as $$6(5-2x)^{-1}$$ and after the chain rule and the power rule we get $$-6(5-2x)^{-2}\cdot (-2)$$ where we have $(x^n))'=nx^{n-1}$

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You can also tackle this using the quotient rule: $$ \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)*g(x)-g'(x)*f(x)}{[g(x)]^2} $$ setting $f(x)=6$, $g(x)=5-2x$, you get $$ \frac{12}{(5-2x)^2} $$ as your derivative.