- #1

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**Homework Statement**

y=5

^{-1/x}

**The attempt at a solution**

y'=5

^{-1/x}ln5(-1)(x

^{-2}/(x^2)

y'=-5

^{-1/x}x

^{-2}ln5/x

^{2}

But apparently my book says that the x^-2 should be -x^-2

That's what I don't understand. Some help please?

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- Thread starter quicksilver123
- Start date

- #1

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y=5

y'=5

y'=-5

But apparently my book says that the x^-2 should be -x^-2

That's what I don't understand. Some help please?

- #2

Mark44

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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}##

You have a mistake above -- that xHomework Statement

y=5^{-1/x}

The attempt at a solution

y'=5^{-1/x}ln5(-1)(x^{-2}/(x^2)

You have the minus sign in -1. The book apparently shows it with -xquicksilver123 said:y'=-5^{-1/x}x^{-2}ln5/x^{2}

But apparently my book says that the x^-2 should be -x^-2

That's what I don't understand. Some help please?

- #3

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The -5 is supposed to be there, but IN ADDITION there is a negative in front of the x in the solution.

- #4

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Solution from the book:

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

y'=5

- #5

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Wolfram verifies the result from the book.

- #6

Ray Vickson

Science Advisor

Homework Helper

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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##.

- #7

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Because without the second negative sign to cancel out the first (the first being -1), my final answer is ending up as negative...

- #8

Mark44

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##y = 5^{-1/x} = (e^{\ln(5)})^{-1/x} = e^{-\ln(5)/x}##

Because without the second negative sign to cancel out the first (the first being -1), my final answer is ending up as negative...

##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?

- #9

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

- #10

Mark44

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

Last edited:

- #11

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

- #12

Mark44

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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.

- #13

Mark44

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@quicksilver123, can you finish this problem?

- #14

SammyS

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No. We don't give step-by-step.Thanks for your reply. Could I please get a step by step method that doesn't use Euler's number

... but try it this way:

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

- #15

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

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