Find from first principle, the derivative of Find from first principle, the derivative of
$$f(x)=\dfrac {ax+b}{\sqrt {x}}.$$
My Attempt:
$$f(x)=\dfrac {ax+b}{\sqrt {x}}$$
$$f(x+\Delta x)=\dfrac {a(x+\Delta x)+b}{\sqrt {x+\Delta x}}$$
where $\Delta x$ is a small increment in $x$.
From first principle,
$$f'(x)=\lim_{\Delta x\to 0} \dfrac {f(x+\Delta x)-f(x)}{\Delta x}$$
$$=\frac {\left(\frac {ax+a\Delta x+b}{\sqrt {x+\Delta x} - \frac {ax+b}{\sqrt {x}}}\right)}{\Delta x}$$
 A: First, simplify the expression you're trying to differentiate into:
$$f(x) = ax^\frac12 + bx^\frac{-1}2.$$
It becomes simple to apply the definition of the derivative:
$$\frac d{dx}f(x) = \lim_{h\to0}\left[\frac{f(x+h)-f(x)}{h}\right]\\
= \lim_{h\to0}\left[\frac{a(x+h)^\frac12+b(x+h)^\frac{-1}2 - \left(ax^\frac12+bx^\frac{-1}2\right)}{h}\right]\\
=\lim_{h\to0}\left[\frac{a(x+h)^\frac12-ax^\frac12}h\right] + \lim_{h\to0}\left[\frac{b(x+h)^\frac{-1}2-bx^\frac{-1}2}h\right]$$
Can you do the rest?
A: Let $t=x+\Delta x$. $$f'(x)=\lim_{\Delta x\to 0} \dfrac {f(x+\Delta x)-f(x)}{\Delta x}$$
$$=\lim_{t\to x} \dfrac {f(t)-f(x)}{t-x}$$
$$=\lim_{t\to x} \dfrac {\frac{at+b}{\sqrt{t}}-\frac{ax+b}{\sqrt{x}}}{t-x}=\lim_{t\to x} \frac{\sqrt x(at+b)-\sqrt t(ax+b)}{\sqrt{t}\sqrt x(t-x)}.$$
Here adding and subtracting the term $\sqrt x(ax+b)$ we get
$$=\lim_{t\to x} \frac{a\sqrt x(t-x)+(ax+b)(\sqrt x-\sqrt t)}{\sqrt{x}\sqrt t (t-x)}$$
$$=\lim_{t\to x} \bigg(\frac{a\sqrt x}{\sqrt x\sqrt t}-\frac{ax+b}{\sqrt{x}\sqrt t(\sqrt{x}+\sqrt t)}\bigg)=\frac{a}{\sqrt x}-\frac{ax+b}{2x\sqrt{x}}=\frac{a}{2\sqrt x}-\frac{b}{2x\sqrt{x}}$$
provided that $x,\,t>0.$
