# Power Rule

Homework Statement
y=5-1/x

The attempt at a solution
y'=5-1/xln5(-1)(x-2/(x^2)
y'=-5-1/xx-2ln5/x2

But apparently my book says that the x^-2 should be -x^-2
That's what I don't understand. Some help please?

## Answers and Replies

Mark44
Mentor
Your thread title, "Power Rule," doesn't apply to the function in this problem, which is in fact an exponential function.
Power rule: ##\frac d {dx}(x^n) = nx^{n - 1}##
Chain rule form of power rule: ##\frac d {dx}(u^n) = nu^{n - 1} \cdot \frac {du}{dx}##
Exponential function rule with chain rule: ##\frac d {dx}(e^u) = e^u \cdot \frac {du}{dx}##
Homework Statement
y=5-1/x

The attempt at a solution
y'=5-1/xln5(-1)(x-2/(x^2)
You have a mistake above -- that x-2 factor shouldn't be there.
quicksilver123 said:
y'=-5-1/xx-2ln5/x2

But apparently my book says that the x^-2 should be -x^-2
That's what I don't understand. Some help please?
You have the minus sign in -1. The book apparently shows it with -x-2. It's like -2 * 5 versus 2 * (-5) -- same result.

No, you misunderstand. I typed what I meant, minus the title error.
The -5 is supposed to be there, but IN ADDITION there is a negative in front of the x in the solution.

Solution from the book:

y'=5-1/x(ln5)[-1*(-x-2)]

Wolfram verifies the result from the book.

Ray Vickson
Homework Helper
Dearly Missed
Solution from the book:

y'=5-1/x(ln5)[-1*(-x-2)]

The book's solution is correct (although I would have written ##1/x^2## or ##x^{-2}## instead of ##(-1)(-x^{-2})##. Just use the fact that for constants ##c## and ##n## we have
$$\frac{d}{dx} c x^n = c n x^{n-1}.$$
Apply this to ##c=-\ln(5)## and ##n = -1##.

I just don't understand where the second negative comes from. I understand the negative one as that's the sign of the exponent, but what's the negative doing on x^-2? (-x^-2)
Because without the second negative sign to cancel out the first (the first being -1), my final answer is ending up as negative...

Mark44
Mentor
I just don't understand where the second negative comes from. I understand the negative one as that's the sign of the exponent, but what's the negative doing on x^-2? (-x^-2)
Because without the second negative sign to cancel out the first (the first being -1), my final answer is ending up as negative...
##y = 5^{-1/x} = (e^{\ln(5)})^{-1/x} = e^{-\ln(5)/x}##
##y' = e^{-\ln(5)/x} \cdot \frac d {dx}(-\ln(5)/x) = 5^{-1/x} \cdot \frac d {dx}(-\ln(5)/x)##

What do you get for the part I didn't finish?

quicksilver123
I dunno I haven't got to derivatives of logs in my book yet :(

Mark44
Mentor
I dunno I haven't got to derivatives of logs in my book yet :(
-ln(5) is a constant that multiplies 1/x. Surely you've seen a rule on how to differentiate a constant times a function.

Last edited:
quicksilver123
Thanks for your reply. Could I please get a step by step method that doesn't use Euler's number

Mark44
Mentor
I dunno I haven't got to derivatives of logs in my book yet :(

-ln(5) is a constant that multiplies 1/x. Surely you've seen a rule on how to differentiate a constant times a function.

Thanks for your reply. Could I please get a step by step method that doesn't use Euler's number
???
Post #8 is nearly complete, and goes step by step. All you have to do is differentiate -ln(5)/x. The last part of the equation I showed in post #8 doesn't involve e.

quicksilver123
SammyS
Staff Emeritus
Homework Helper
Gold Member
Thanks for your reply. Could I please get a step by step method that doesn't use Euler's number
No. We don't give step-by-step.

... but try it this way:

Start with ##\ \displaystyle y=\left(\frac{1}{5} \right)^{1/x} \,.\ ##

I've finished that problem I appreciate it. It seems I've been making dumb mistakes with the signs.