# Finding a derivative by definition

• Nov 10th 2009, 07:14 AM
Em Yeu Anh
Finding a derivative by definition
(Happy) First of all, hi everyone, I'm new to this forum

I'll cut to the chase. Q: Let $\displaystyle f(x) = \frac{1}{\sqrt{x}} + 2x^2, x>0$. Find f'(x) using the definition of the derivative.

$\displaystyle f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h}= \lim_{h\to0}\frac{\frac{1}{\sqrt{x+h}}+2(x+h)^2 -\frac{1}{\sqrt{x}}-2x^2}{h}$ Clueless on how to simplify this.
• Nov 10th 2009, 08:09 AM
stapel
Are you allowed to break this apart into two limits?

You've got f(x) = g(x) + h(x), for g(x) = 1/sqrt[x] and h(x) = 2x^2. The h(x) limit is easy, and the g(x) limit isn't that hard, once it's pulled aside and you apply the conjugate (a trick you'll want to keep in mind for the next test):

$\displaystyle \frac{g(x\, +\, h)\, -\, g(x)}{h}\, =\, \frac{\frac{1}{\sqrt{x\, +\, h}}\, -\, \frac{1}{\sqrt{x}}}{h}$

$\displaystyle =\, \frac{\left(\frac{\sqrt{x}\, -\, \sqrt{x\, +\, h}}{\sqrt{x}\sqrt{x\, +\, h}}\right)}{h}\, =\, \frac{\sqrt{x}\, -\, \sqrt{x\, +\, h}}{\sqrt{x}\sqrt{x\, +\, h}(h)}$

$\displaystyle =\, \left(\frac{\sqrt{x}\, -\, \sqrt{x\, +\, h}}{\sqrt{x}\sqrt{x\, +\, h}(h)}\right)\left(\frac{\sqrt{x}\, +\, \sqrt{x\, +\, h}}{\sqrt{x}\, +\, \sqrt{x\, +\, h}}\right)$

$\displaystyle =\, \frac{x\, -\, (x\, +\, h)}{\sqrt{x}\sqrt{x\, +\, h}(h)\left(\sqrt{x}\, +\, \sqrt{x\, +\, h}\right)}$

$\displaystyle =\, \frac{-1}{\sqrt{x}\sqrt{x\, +\, h}\left(\sqrt{x}\, +\, \sqrt{x\, +\, h}\right)}$

Now take the limit. (Wink)