$$y=5^{-1/x}$$
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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$\begingroup$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).
$\endgroup$ 2 $\begingroup$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 $$y = 5^{-\frac{1}{x}}$$ $$y' = 5^{-\frac{1}{x}} \cdot \ln(5) \cdot \frac{1}{x^2}$$
$\endgroup$ 2 $\begingroup$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 $x$.
$\frac{d}{dx} = 5^{-1/x}ln(5)*1/x^2$
$\endgroup$ $\begingroup$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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