f(x)=(x^5-4/x)^-4

f'(x)=

gave me -4(5x^4-4)/(x^5-4/x)^5 , but its wrong

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- Apr 19th 2010, 06:13 PMwannousderivative
f(x)=(x^5-4/x)^-4

f'(x)=

gave me -4(5x^4-4)/(x^5-4/x)^5 , but its wrong - Apr 19th 2010, 06:16 PMharish21
- Apr 19th 2010, 06:29 PMwannous
only the -4 is divided by x

- Apr 19th 2010, 06:35 PMdrumist
So it is:

$\displaystyle f(x)=\left( x^5 - \frac{4}{x} \right) ^{-4} $

It looks like your derivative of $\displaystyle x^5 - \frac{4}{x}$ was wrong.

$\displaystyle \frac{d}{dx} \left( x^5 - \frac{4}{x}\right) = 5x^4 + \frac{4}{x^2}$ - Apr 19th 2010, 06:41 PMwannous
aww so i needed to use quotient rule on the 4/x

- Apr 19th 2010, 06:49 PMdrumist
That does work, but you can also do it by rewriting it first as $\displaystyle 4x^{-1}$ then using the power rule, which gives you $\displaystyle -4x^{-2}$, then changing it back to a fraction to get $\displaystyle -\frac{4}{x^2}$.

By the way, in general, you'll probably want to just commit to memory that the derivative of $\displaystyle \frac{1}{x}$ is $\displaystyle -\frac{1}{x^2}$ since it seems to come up often enough that it will be a time saver to just learn it.