Why the derivative of $y = \frac{x^5}{a+b}-\frac{x^2}{a-b}-x$ is solved by deriving just the numerators?
The solution is $\frac{dy}{dx}=\frac{5x^4}{a-b}-\frac{2x}{a-b}-1$.
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As $$\dfrac{d[a\cdot f(x)]}{dx}=a\cdot \dfrac{d[ f(x)]}{dx}$$ where $a$ is independent of $x$ – lab bhattacharjee Apr 03 '15 at 14:50
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1Note: $x$ is different from $a,b$ because you are taking $\frac{d}{dx}$, the derivative with respect to $x$. – vadim123 Apr 03 '15 at 14:58
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To whom it may concern: please don't downvote questions because you think they're elementary. – Jack M Apr 03 '15 at 14:58
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$a, b$ are just constants. it like differentiating $\frac{x^5}{2} - \frac {x^2}3 - x$ – abel Apr 03 '15 at 15:03
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Because the denominator does not depend on $x$. So if we were to formally use $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$ we get by plugging $v'=0$ ($v$ does not depend on $x$) $\left(\frac{u}{v}\right)'=\frac{u'}{v}$
marwalix
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But this would be a division by zero. That's why we ignore the constants in the denominator? – juliano.net Apr 03 '15 at 16:18
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If the denominator is constant, it's really easiest to apply the constant multiple rule. That is, $\frac{d}{dx}\frac{f(x)}{c}=\frac{d}{dx}\frac{1}{c}f(x)=\frac1c f'(x)=\frac{f'(x)}{c}$
If you insist on using the quotient rule, you get the same answer, as shown in marwalix's post.
paw88789
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