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Help would be so greatly appreciated :] It's another homework problem...I unfortunately was not present during the lecture for these types of problems. I'm guessing from the $-1/x$ there would be a $\ln()$ in the answer?

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Is that $5^{-1/x}$, or $\frac{5^{-1}}x$? (I’d just about bet on the former, but let’s make sure.) – Brian M. Scott Feb 20 '13 at 3:09
definitely the first one :] – bkaifos15 Feb 20 '13 at 3:10
I bet the goal is to find the derivative of a function, not the derivative of a question. – Jonas Meyer Feb 20 '13 at 3:10
There's examples on how to do this here just for possible reference – TakeS Feb 20 '13 at 3:11
Jonas: Too bad I already asked that question (which is why I said of this question not this function). Thanks. – bkaifos15 Feb 20 '13 at 3:13
up vote 1 down vote accepted

It’s of the form $a^u$, where $u$ is some function of $x$. One of the basic differentiation formulas is

$$\frac{d}{dx}a^x=a^x\ln a\;;$$

combine that with the chain rule, since the exponent isn’t just $x$, and you get

$$\frac{dy}{dx}=5^{-1/x}(\ln 5)\left[-\frac1x\right]'\;.$$

Now you need the derivative of $-\frac1x$. Write it as $-x^{-1}$, and you see that all you need is the power rule:

$$\frac{dy}{dx}=5^{-1/x}(\ln 5)\left[-\frac1x\right]'=5^{-1/x}(\ln 5)x^{-2}=\frac{\ln 5}{x^25^{1/x}}$$

(among many possible final forms).

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Thanks again! Definitely Helpful! – bkaifos15 Feb 20 '13 at 3:39
@bkaifos15: My pleasure. – Brian M. Scott Feb 20 '13 at 3:39

Notice that $$ y = 5^{- \frac{1}{x}} = \exp \left[ - \frac{1}{x} \cdot \ln(5) \right]. $$ Therefore, \begin{align} \frac{dy}{dx} &= \exp \left[ - \frac{1}{x} \cdot \ln(5) \right] \cdot \frac{d}{dx} \left[ - \frac{1}{x} \cdot \ln(5) \right] \quad (\text{By the Exponential Chain Rule.}) \\ &= 5^{- \frac{1}{x}} \cdot \frac{d}{dx} \left[ - \frac{1}{x} \cdot \ln(5) \right] \\ &= 5^{- \frac{1}{x}} \cdot \frac{1}{x^{2}} \cdot \ln(5) \\ &= \frac{\ln(5)}{5^{\frac{1}{x}} x^{2}}. \end{align}

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This has the form $a^{g(x)}$, where $a$ is a constant and $g(x)$ is some function of $x$. $$\frac{d}{dx}(a^{g(x)})=a^{g(x)} \cdot \ln(a) \cdot g'(x)$$

Don't forget to take the derivative of the exponent $g(x)$ by the chain rule.

You have $$y = 5^{-\frac{1}{x}}$$ $$y' = 5^{-\frac{1}{x}} \cdot \ln(5) \cdot \frac{1}{x^2}$$

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This is a little off. First of all that first equation is wrong $\frac{d}{dx}(a^x) = a^x ln(a)$. There is no $dx$ term at the end. Secondly your final answer includes ln(x). It should be ln(5). – daniel.wright Feb 20 '13 at 3:20
Thanks. Fixed it. – badjr Feb 20 '13 at 3:21
What you mean to say is that $\frac{d}{dx} a^{g(x)} = a^{g(x)}ln(a)*g'(x)$. – daniel.wright Feb 20 '13 at 3:22

Use the chain rule. So start by differentiating $5^{-1/x}$ with respect to $-1/x$ then multiple by the derivative of $-1/x$ with respect to $x$.

$\frac{d}{dx} = 5^{-1/x}ln(5)*1/x^2$

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