Question

derivative of (x^3)/2 + 1/6x?

Answer 1

\[
f(x)=\frac{x^3}{2}+\frac{1}{6x}\\
f(x)=\frac{x^3}{2}+\frac{1}{6}x^{-1}\\
f'(x)\frac{3x^2}{2}-\frac{1}{6}x^{-2}
\]
Just uses the Power Rule and the fact that (f+g)'(x)=f'(x)+g'(x)

Answer 2

Unless by 1/6x you meant \(\dfrac{1}{6}x\), in which case the second term would just be \(\dfrac{1}{6}\).

Answer 3

yes, i did mean 1/(6x)

Answer 4

hang on, still working out the solution to make sure i get it right..

Answer 5

what is that second rule? supposed to be quotient rule?

Answer 6

No need for quotient rule. The second rule just means that you can take the derivatives of terms separately if they are added together. Dunno if there's a formal name for it.

Answer 7

power rule only applies to the second fraction, right? wouldn't quotient have to be applied to the first fraction to find its derivative?

Answer 8

The power rule can work on them both, because the power rule is general enough to apply to both positive and negative exponents.

Answer 9

You could use the quotient rule and obtain the same result, it's just overkill.

Answer 10

i got the answer now. what happens to the 2 in the denominator of the first fraction when applying product rule?

Answer 11

You don't need to use the product rule either, just the power rule. The 2 stays there.

Answer 12

ah, ok. sorry, i meant power rule. so constant stays same in fraction when finding derivative?

Answer 13

Power rule just says that the derivative of \(x^n\) is \(nx^{n-1}\). So, if you have \(\frac{1}{2}x^n\), the derivative is still \(\frac{1}{2}nx^{n-1}\), the coefficient doesn't go away.

Answer 14

The proof of that is actually just a special application of the product rule :)

Answer 15

Say you have \(f(x)=cx^n\), you can define \(g(x)=c\) and \(h(x)=x^n\), and then \(f(x)=g(x)\cdot h(x)\). If you apply the product rule, you'll see that you can prove for yourself that \(f'(x)=cnx^{n-1}\).

Answer 16

thanks, nbouscal. 'preciate it :-)

Answer 17

Any time :)