Calculating the derivative of ${10^x-1 \over x}$ using the limit definition

Question

I've been stuck for the last couple of hours trying to calculate the derivative of $$f(x) ={10^x-1 \over x}$$ With the limit definition of the derivative: $$\lim \limits_{h \to 0} \frac{f(x+h) -f(x)}{h}$$ I just can't get rid of $h$, I think I'm missing an important formula here.

Answer 1

Follow the limit definition of derivative $$\begin{align} f'(x) &= \lim_{h\to 0} \frac{\frac{10^{x+h}-1}{x+h}-\frac{10^{x}-1}{x}}{h} \\ &= \lim_{h\to 0} \frac{10^x(x(10^h-1)-h)+h}{h} \cdot \frac1{x(x+h))} \end{align}$$

Now you can split limit over sum and use $\lim\limits_{h\to 0} \frac{a^h - 1}{h} = \ln a$ to get:

$$\dfrac{10^x (x\ln(10)-1)+1}{x^2}$$

Answer 2

$\frac{f(x+h)-f(x)}{h}=\frac{\frac{10^{x+h}-1}{x+h}-\frac{10^{x}-1}{x}}{h}= \frac{x10^{x}(10^h-1)+h-h10^{x}}{x(x+h)h}=\frac{1-10^{x}}{x(x+h)}+\frac{x10^{x}}{x(x+h)}\frac{10^h-1}{h}$.

You know first term converges to $\frac{1-10^x}{x^2}$. For the second term use the same argument as for first part and l'hopital's rule for $\frac{10^h-1}{h}$.As both parts have finite limit, the limit of the product is the product of the limits.