Understanding y'=5-1/xln5(-1)(x-2/x^-2)

[quicksilver123]

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?

 

[Mark44]

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.

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.

 

[quicksilver123]

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.

 

[quicksilver123]

Solution from the book:

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

 

[quicksilver123]

Wolfram verifies the result from the book.

 

[Ray Vickson]

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 $n = -1$.

 

[quicksilver123]

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]

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

 

[Mark44]

I don't know 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.

 

[quicksilver123]

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

 

[Mark44]

I don't know 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.

 

[Mark44]

@quicksilver123, can you finish this problem?

 

[SammyS]

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} \,.\$

 

[quicksilver123]

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