Derivative using first principle method

Question

How to find the derivative of $\frac{(x+1)}{\sqrt x}$ using first principle method?

I am able to solve the above using binomial theorem but i want to learn the method without binomial theorem.

Answer 1

The derivative of a fraction $f(x)/g(x)$ is given by $$\frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2}.$$ Where, in your case, $f(x) = x+1$, $g(x) = \sqrt{x}$, $f'(x) =1$, $g'(x) = 1/(2\sqrt x)$.

Answer 2

You can get away with it by observing that $$ f(x)=\sqrt{x}+\frac{1}{\sqrt{x}} $$ Then you can do, for $x>0$, \begin{align} \lim_{h\to0} \frac{\sqrt{x+h}+\dfrac{1}{\sqrt{x+h}}-\sqrt{x}-\dfrac{1}{\sqrt{x}}}{h} &=\lim_{h\to0}\left( \frac{\sqrt{x+h}-\sqrt{x}}{h}+ \frac{\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt{x}}}{h} \right) \\[6px] &=\lim_{h\to0}\left( \frac{1}{\sqrt{x+h}+\sqrt{x}}+ \frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\,\sqrt{x+h}} \right) \end{align} and you can finish it up easily.

In other words, pretend you're ignoring that the derivative of the sum is the sum of the derivatives, but act with this important knowledge in the background.

Answer 3

By definition;

$f'(x) =\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

$f(x)= \dfrac{x+1}{\sqrt x}$

$f(x+h) = \dfrac{x+h+1}{\sqrt{x+h}}$

$f'(x) = \lim_{h \to0}\dfrac{\frac{x+h+1}{\sqrt{x+h}}-\frac{x+1}{\sqrt{x}}}{h}$

$f'(x)= \lim_{h\to0}\dfrac{\sqrt{x}(x+h+1)-\sqrt{x+h}(x+1)}{\sqrt{x}\cdot\sqrt{x+h}\cdot h}$

$f'(x) =\lim_{h\to0}\dfrac{\sqrt{x}(x+h+1)-\sqrt{x+h}(x+1)}{\sqrt{x}\cdot\sqrt{x+h}\cdot h}\times \dfrac{\sqrt{x}(x+h+1)+\sqrt{x+h}(x+1)}{\sqrt{x}(x+h+1)+\sqrt{x+h}(x+1)}$

$f'(x) =\lim_{h\to0}\dfrac{x(x+h+1)^2-(x+h)(x+1)^2}{x\cdot h\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot h\cdot \sqrt{x}(x+1)}$

$f'(x) =\lim_{h\to0}\dfrac{x(x+h+1)^2-x(x+1)^2}{x\cdot h\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot h\cdot \sqrt{x}(x+1)}-\dfrac{h(x+1)^2}{x\cdot h\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot h\cdot \sqrt{x}(x+1)}$

$f'(x) =\lim_{h\to0}\dfrac{x(x+h+1)^2-x(x+1)^2}{x\cdot h\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot h\cdot \sqrt{x}(x+1)}-\dfrac{(x+1)^2}{x\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot \sqrt{x}(x+1)}$

expand the numerator of the first term gives ;

$f'(x) = \lim_{h\to0}\dfrac{ x(x^2+2xh+2x+h^2+2h+1)-x(x^2+1+2x)}{x\cdot h\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot h\cdot \sqrt{x}(x+1)}-\dfrac{(x+1)^2}{x\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot \sqrt{x}(x+1)}$

$f'(x) =\lim_{h\to0}\dfrac{2x^2h+h^2x+2hx}{x\cdot h\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot h\cdot \sqrt{x}(x+1)}-\dfrac{(x+1)^2}{x\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot \sqrt{x}(x+1)}$

$f'(x) =\lim_{h\to0}\dfrac{2x^2+hx+2x}{x\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot \sqrt{x}(x+1)}-\dfrac{(x+1)^2}{x\cdot\sqrt{x+h}(x+h+1)+(x+h)\cdot \sqrt{x}(x+1)}$

applying the limit gives;

$f'(x)= \dfrac{2x(x+1)}{2x\sqrt{x}(x+1)}-\dfrac{(x+1)^2}{2x(x+1)\sqrt{x}}$

$f'(x)= \dfrac1{\sqrt x}-\dfrac{x}{2x\sqrt{x}}-\dfrac1{2x\sqrt{x}}$

$f'(x) =\dfrac{1}{\sqrt x}-\dfrac{1}{2\sqrt{x}}-\dfrac1{2x\sqrt{x}}$

$\therefore f'(x)=\frac 12 x^{-\frac12}-\frac12x^{-\frac 32}$