Give the interval for which $\frac{\ln(t)}{1+t^{x}}$ is differentiable 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.
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 $t&gt;0$,

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

Thus $f'(x) = \frac{-t^x \ln(t)^2}{(1+t^x)^2}$
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 $t&gt;0$,

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

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