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We are given the function: $$f(x) = \frac{\ln(t)}{1+t^{x}}$$ and are asked to show for which interval $I \subset \Bbb{R}$, where I can be unbounded, $f(x)$ is derivable.

I'm not sure how to start working this out, so any help would be appreciated. Thank you in advance.

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    You have $\frac{1}{1+t^{x}}$ in the title and $\frac{\ln(t)}{1+t^{x}}$ in the body. Which one do you want to know is driveable? – 2017-01-29
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    Do you know the chain rule and the derivative of $t^x$? – 2017-01-29
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    Sorry for the confusion, I edited the title a bit to make it more clear. And yes we know the chain rule, but not the derivative of $t^{x}$ if we don't prove it. – 2017-01-30

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I assume you mean differentiable.

Since it's a function of $x$, the factor $\ln(t)$ can be viewed as a constant, just use the linearity of the derivative in the end. Since nothing is mentioned I assumed that $f: \mathbb{R} \to \mathbb{R}$ and then $t>0$,

$$ \lim_{h \to 0}\frac{\frac{1}{1+t^{x+h}} - \frac{1}{1+t^x}}{h} = \lim_{h \to 0} \frac{1+t^x - (1 + t^{x+h})}{h(1+t^x)(1+t^{x+h})} = \\ \lim_{h \to 0} \frac{t^x(1-t^{h})}{h(1+t^x)(1+t^{x+h})} = \frac{t^x}{1+t^x} \cdot \lim_{h \to 0}\frac{1-t^h}{h(1+t^{x+h})} =\\ / \text{ h'opitals rule }/ = \frac{t^x}{1+t^x} \cdot \lim_{h \to 0}\frac{-t^h \ln(t)}{t^{h+x} + 1 + h\ln(t)t^{h+x}} \\ = \frac{-t^x \ln(t)}{(1+t^x)^2} $$

Thus $f'(x) = \frac{-t^x \ln(t)^2}{(1+t^x)^2}$

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    Why downvotes? If I have made mistakes, please let me know. – 2017-01-29
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    I did not downvote, but I assume they are because you didn't really answer the question. The OP asked the interval where $f(x)$ is derivable, not the derivative of the function. Also, the calculation of the derivative might be a bit overkill, since you don't really have to use the definition (but it is very nice and interesting calculation thou). – 2017-01-29
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    The title of the question says "Show that blabla is deriveable". And to answer the WHERE we need to know where the limit exists. So my train of thought was that if I showed that it's Differentiable, the op can reason when the limits exists. But I understand your point, thanks! @mitchbus – 2017-01-29
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    Sorry, I edited the title to make the question more clear. – 2017-01-30
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    For which points in $\mathbb{R}$ is the above calculation valid? @Pyrus – 2017-02-06