Find the Derivative $ 7^{\ln x} $ using first principle I still can't figure this out,
Question is Find the Derivative $ 7^{\ln(x)} $ using first principle 
This is where I got
$$\lim_{h \to 0} \frac{7^{\ln(x+h)} -  7^{\ln(x)}}{h}  $$
then What should I do?
 A: Note that calculating derivative of $f(x)$ via first principles means that we need to calculate the limit $$\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$ without using any rules of differentiation.
Here $f(x) = 7^{\log x}$ and we can proceed as follows
\begin{align}
f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\
&= \lim_{h \to 0}\frac{7^{\log(x + h)} - 7^{\log x}}{h}\notag\\
&= 7^{\log x}\lim_{h \to 0}\frac{7^{\log(x + h) - \log x} - 1}{h}\notag\\
&= 7^{\log x}\lim_{h \to 0}\frac{7^{\log(x + h) - \log x} - 1}{\log(x + h) - \log x}\cdot\frac{\log(x + h) - \log x}{h}\notag\\
&= 7^{\log x}\lim_{h \to 0}\frac{7^{\log(x + h) - \log x} - 1}{\log(x + h) - \log x}\cdot\lim_{h \to 0}\frac{\log(x + h) - \log x}{h}\notag\\
&= 7^{\log x}\lim_{t \to 0}\frac{7^{t} - 1}{t}\cdot\lim_{h \to 0}\frac{\log((x + h)/x)}{h}\text{ (by putting }t = \log(x + h) - \log x)\notag\\
&= 7^{\log x}\lim_{t \to 0}\frac{e^{t\log 7} - 1}{t}\cdot\lim_{h \to 0}\frac{\log(1 + (h/x))}{h}\notag\\
&= 7^{\log x}\lim_{t \to 0}\log 7 \cdot\frac{e^{t\log 7} - 1}{t\log 7}\cdot\lim_{h \to 0}\frac{\log(1 + (h/x))}{h/x}\cdot\frac{1}{x}\notag\\
&= \frac{7^{\log x}\log 7}{x}\lim_{y \to 0}\cdot\frac{e^{y} - 1}{y}\cdot\lim_{z \to 0}\frac{\log(1 + z)}{z}\text{ (putting }y = t\log 7, z = h/x)\notag\\
&= \frac{7^{\log x}\log 7}{x}\cdot 1\cdot 1\notag\\
&= \frac{7^{\log x}\log 7}{x}
\end{align}
A: Note that $7^{\ln(x)} = \left(e^{\ln(7)} \right)^{\ln(x)} = \left(e^{\ln(x)} \right)^{\ln(7)} = x^{\ln(7)}$.
From the generalized binomial theorem, we have
$$(x+h)^{\alpha} = \sum_{k=0}^{\infty} \dbinom{\alpha}k x^{\alpha-k}h^k = x^{\alpha} + \alpha x^{\alpha-1}h + h^2 f(x,h;\alpha)$$
where $f(x,0;\alpha)$ is continuous in $h$ with $f(x,0;\alpha) = 0$. Hence,
$$\dfrac{(x+h)^{\alpha}-x^{\alpha}}h = \alpha x^{\alpha-1} + hf(x,h;\alpha)$$
Hence, we have
$$\lim_{h \to 0}\dfrac{(x+h)^{\alpha}-x^{\alpha}}h = \alpha x^{\alpha-1} + \lim_{h \to 0}hf(x,h;\alpha) = \alpha x^{\alpha-1}$$
Hence, we have
$$\lim_{h \to 0} \dfrac{7^{\ln(x+h)}-7^{\ln(x)}}h = \lim_{h \to 0} \dfrac{(x+h)^{\ln(7)}-x^{\ln(7)}}h = \ln(7)x^{\left(\ln(7)-1\right)}$$
A: I'm not sure if this is "first principle" or not, but since $7^a = e^{(\ln x)(\ln 7)}$ for any real $a$ you have $$7^{\ln x} = e^{(\ln x)(\ln 7)}.$$ Thus $$\lim_{h \to 0} \frac{7^{\ln(x+h)} - 7^{\ln x}}{h} = \frac d{dx} e^{(\ln x)(\ln 7)}$$ which you can compute using the usual rules of differentiation.
A: This is the derivative of $(7)^{\ln x}$ or the derivative of $e^{\ln x\cdot  \ln 7}$ or the derivative of $x^{\ln 7}$ which is $\ln 7 \cdot x^{(\ln 7) -1}=(1/x)\ln7\cdot x^{\ln7}$.
A: $$\lim_{h \to 0} \frac{7^{\ln(x+h)}-7^{\ln(x)}}{h}=$$
$$7^{\ln(x)}\lim_{h \to 0} \frac{7^{\ln(x+h)-\ln(x)}-7^{\ln(x)-\ln(x)}}{h}$$
$$7^{\ln(x)}\lim_{h \to 0} \frac{7^{\ln(1+h/x)}-1}{h}$$
I use the Maclaurin series expansion of $\ln(1+x)=x-x^2/2+x^3/3-x^4/4+\,\dots$: 
$$7^{\ln(x)}\lim_{h \to 0} \frac{7^{h/x}-1}{h}$$
$$7^{\ln(x)}\lim_{h \to 0} \frac{(7^{1/x})^h-1}{h}$$
Now I use the well-known limit $\displaystyle\lim_{h \to 0}\frac{a^h-1}{h}=\ln(a)$:
$$7^{\ln(x)}\ln(7^{1/x})=\frac{7^{\ln(x)}\ln(7)}{x}$$
