$f(x) = x+\sqrt{x}+4$
Obviously the derivative is $1 +\frac{1}{2\sqrt{x}}$
My problem is with using the precise definition of a limit namely:
$\frac{d}{{dx}}f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
How would you solve this using the precice definition of a derivative?
$\endgroup$ 11 Answer
$\begingroup$\begin{align*}\frac{f(x+h)-f(x)}{h} &= \frac{(x+h)+\sqrt{x+h}+4-(x+\sqrt{x}+4)}{h} \\ &= \frac{h + \sqrt{x+h}-\sqrt{x}}{h} \\ &= 1+\frac{\sqrt{x+h}-\sqrt{x}}{h} \\ &= 1 + \frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\\ &= 1 + \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\ &= 1 + \frac{1}{\sqrt{x+h}+\sqrt{x}} \end{align*}
$\endgroup$ 1
