# How do you use the definition of the derivative to find f'(x) if f(x)=1/sqrtx?

Sep 28, 2016

Start with lim/(h→0){f(x+h)-f(x)}/h multiply numerator and denominator by $f \left(x + h\right) + f \left(x\right)$

#### Explanation:

lim/(h→0){(f(x+h))^2-(f(x))^2}/(h(f(x) + f(x+h)))

We observe that ${\left(f \left(x\right)\right)}^{2} = \frac{1}{x}$ and ${\left(f \left(x + h\right)\right)}^{2} = \frac{1}{x + h}$ and we do the substitution:

lim/(h→0){1/(x+h) - 1/x}/(h(1/sqrt(x) + 1/(sqrt(x+h)))

Make a common denominator for $\frac{1}{x + h} - \frac{1}{x} = \frac{x}{\left(x \left(x + h\right)\right)} - \frac{x + h}{\left(x \left(x + h\right)\right)} = - \frac{h}{\left(x \left(x + h\right)\right)}$

Substitute back into the limit:

lim/(h→0){-h/((x(x +h)))}/(h(1/sqrt(x) + 1/(sqrt(x+h)))

Please observe that $- \frac{h}{h} = - 1$

lim/(h→0){-1/((x(x +h)))}/((1/sqrt(x) + 1/(sqrt(x+h)))

Now it is safe to let h go to zero:

$\frac{- \frac{1}{\left(x \left(x\right)\right)}}{\left(\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}}\right)} = \frac{- \frac{1}{x} ^ 2}{\frac{2}{\sqrt{x}}} = - \frac{1}{2 {x}^{\frac{3}{2}}}$