# Problem using the difference quotient

• July 17th 2012, 08:29 AM
astartleddeer
Problem using the difference quotient
Hello,

I'm trying to differentiate a sum of functions by means of the difference quotient from a question in my book. The derivative is a given, but I can't end up at the final result (I can, but only if I use $anx^{n -1}$)

I feel like I'm close but I now need assistance.

$f(x) = \frac{7}{2} x^\frac{1}{2} + \frac{3}{x^2} - 9$

$f(x + \Delta x) = \frac{7}{2} (x + \Delta x)^\frac{1}{2} + \frac{3}{(x + \Delta x)^2} - 9$

$f(x + \Delta x) - f(x) = \left[\frac{7}{2} (x + \Delta x)^\frac{1}{2} + \frac{3}{(x + \Delta x)^2} - 9\right] - \left[\frac{7}{2} x^\frac{1}{2} + \frac{3}{x^2} - 9\right]$

$\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{\frac{7}{2}\left[(x +\Delta x)^\frac{1}{2} - x^\frac{1}{2}\right] + 3\left[ \frac{1}{(x + \Delta x)^2} - \frac{1}{x^2}\right]}{\Delta x}$

Now this is where I'm stuck. If I assume I can eliminate $\left[(x +\Delta x)^\frac{1}{2} - x^\frac{1}{2}\right]$ by way of the difference of two squares. Then I can multiply both the numerator and denominator by $\left[(x +\Delta x)^\frac{1}{2} + x^\frac{1}{2}\right]$ in order to replace it with $(x + \Delta x) - x$ in the numerator.

As for $\left[\frac{1}{(x + \Delta x)^2} - \frac{1}{x^2}\right]$ again I think I can eliminate the squares to a more easier form to deal with by way of the difference of two squares but don't know how?

• July 17th 2012, 09:54 AM
Reckoner
Re: Problem using the difference quotient
Quote:

Originally Posted by astartleddeer
$\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{\frac{7}{2}\left[(x +\Delta x)^\frac{1}{2} - x^\frac{1}{2}\right] + 3\left[ \frac{1}{(x + \Delta x)^2} - \frac{1}{x^2}\right]}{\Delta x}$

On the left side, rationalize the numerator (this is a strategy that is often helpful in situations like this), and on the right, combine fractions and expand.

$\lim_{\Delta x\to0}\frac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x\to0}\frac{7(x+\Delta x)^{1/2}-7x^{1/2}}{2\Delta x} + \lim_{\Delta x\to0}\frac1{\Delta x}\left[\frac3{(x+\Delta x)^2} - \frac3{x^2}\right]$

$= \frac72\lim_{\Delta x\to0}\frac{(x+\Delta x)^{1/2}-x^{1/2}}{\Delta x} + 3\lim_{\Delta x\to0}\frac{x^2 - (x+\Delta x)^2}{x^2\Delta x(x+\Delta x)^2}$

$= \frac72\lim_{\Delta x\to0}\frac{\left[(x+\Delta x)^{1/2}-x^{1/2}\right]\left[(x+\Delta x)^{1/2}+x^{1/2}\right]}{\Delta x\left[(x+\Delta x)^{1/2}+x^{1/2}\right]} + 3\lim_{\Delta x\to0}\frac{x^2 - \left(x^2+2x\Delta x+\Delta x^2\right)}{x^2\Delta x(x+\Delta x)^2}$

$= \frac72\lim_{\Delta x\to0}\frac{(x+\Delta x)-x}{\Delta x\left[(x+\Delta x)^{1/2}+x^{1/2}\right]} - 3\lim_{\Delta x\to0}\frac{2x\Delta x+\Delta x^2}{x^2\Delta x(x+\Delta x)^2}$

$= \frac72\lim_{\Delta x\to0}\frac{\Delta x}{\Delta x\left[(x+\Delta x)^{1/2}+x^{1/2}\right]} - 3\lim_{\Delta x\to0}\frac{\Delta x\left(2x+\Delta x\right)}{x^2\Delta x(x+\Delta x)^2}$

$= \frac72\lim_{\Delta x\to0}\frac1{(x+\Delta x)^{1/2}+x^{1/2}} - 3\lim_{\Delta x\to0}\frac{2x+\Delta x}{x^2(x+\Delta x)^2}$

$= \frac7{2\left(x^{1/2}+x^{1/2}\right)} - \frac{3(2x)}{x^2\cdot x^2}$

$= \frac7{4\sqrt x} - \frac6{x^3}$
• July 17th 2012, 10:10 AM
astartleddeer
Re: Problem using the difference quotient
Beautiful! Thanks!