Differentiation (First Principles) - Bogged down!

**Stoot** · June 6th 2012, 02:20 PM

Hi,

Differentiate using first principles:

y = 2x^-2 + 3x^-1

I keep getting bogged down at the point:

1/(2x^2 + 4xh + 2h^2) + 1/(3x + 3h)

as I have to subtract (2x^-2 + 3x^-1) from the above.

If anyone could help i would greatly appreciate it.

Thanks

**skeeter** · June 6th 2012, 02:41 PM

$f(x) = \frac{2}{x^2} + \frac{3}{x}$

$f(x+h) = \frac{2}{(x+h)^2} + \frac{3}{x+h}$

$f'(x) = \lim_{h\to 0} \frac{1}{h} \left[{\frac{2}{(x+h)^2} + \frac{3}{x+h} - \frac{2}{x^2} - \frac{3}{x} \right]$

$f'(x) = \lim_{h\to 0} \frac{1}{h} \left[{\frac{2x^2}{x^2(x+h)^2} + \frac{3x^2(x+h)}{x^2(x+h)^2} - \frac{2(x+h)^2}{x^2(x+h)^2} - \frac{3x(x+h)^2}{x^2(x+h)^2} \right]$

$f'(x) = \lim_{h\to 0} \frac{1}{h} \left[\frac{2x^2 + 3x^3 + 3x^2h - 2(x^2+2xh+h^2) - 3x(x^2+2xh+h^2)}{x^2(x+h)^2}\right]$

take it from here?

**Reckoner** · June 6th 2012, 02:49 PM

Originally Posted by Stoot

y = 2x^-2 + 3x^-1

$y = 2x^{-2} + 3x^{-1}$

$\frac{dy}{dx} = \lim_{h\to0}\frac{2(x+h)^{-2}+3(x+h)^{-1}-2x^{-2}-3x^{-1}}h$

$= \lim_{h\to0}\frac1h\left[\frac2{(x+h)^2}+\frac3{x+h}-\frac2{x^2}-\frac3x\right]$

$= \lim_{h\to0}\frac1h\left[\frac{2x^2+3x^2(x+h)-2(x+h)^2-3x(x+h)^2}{x^2(x+h)^2}\right]$

$= \lim_{h\to0}\frac1h\left[\frac{2x^2+3x^3+3hx^2-2x^2-4hx-2h^2-3x^3-6hx^2-3h^2x}{x^2(x+h)^2}\right]$

$= \lim_{h\to0}\frac1h\left[\frac{-3hx^2-4hx-2h^2-3h^2x}{x^2(x+h)^2}\right]$

Factor out the $h$ and reduce.

**Stoot** · June 6th 2012, 03:23 PM

thanks for the replies guys, I follow it to the step above, but I'm still struggling, specifically with how to factor out, the examples I've dealt with so far are nowhere near this level.

Greatly appreciated.

**Reckoner** · June 6th 2012, 03:28 PM

Originally Posted by Stoot

thanks for the replies guys, I follow it to the step above, but I'm still struggling, specifically with how to factor out, the examples I've dealt with so far are nowhere near this level.

Factoring out a common factor is basic algebra. You should already know how to do that...

$\lim_{h\to0}\frac{-3hx^2-4hx-2h^2-3h^2x}{hx^2(x+h)^2}$

$= \lim_{h\to0}\frac{h\left(-3x^2-4x-2h-3hx\right)}{hx^2(x+h)^2}$

Now cancel out the common factor of $h$ . That will allow you to use direct substitution for the limit.

**Stoot** · June 6th 2012, 04:58 PM

I don't know whether it's because it's 2 am or because I'm having a brain melt but I just can't make this work. You're right, I should know how to do this but I have just started a maths course after along time 'away from the game'!!

**Reckoner** · June 6th 2012, 05:19 PM

Originally Posted by Stoot

You're right, I should know how to do this but I have just started a maths course after along time 'away from the game'!!

The material you are learning now and in the future will depend on your knowledge of the prior material. If you've had a significant gap in your learning, then you really should make the effort to give yourself a serious review of your earlier courses. You could read through your old textbooks if you have them, or look for tutorials on the web.

$\lim_{h\to0}\frac{h\left(-3x^2-4x-2h-3hx\right)}{hx^2(x+h)^2}$

$=\lim_{h\to0}\frac{-3x^2-4x-2h-3hx}{x^2(x+h)^2}$

Notice that when $h=0$ , the denominator is no longer zero. Since we're taking the limit of a rational function whose denominator is not zero, you can calculate the value of the limit by substituting $h=0$ into the expression:

$=\lim_{h\to0}\frac{-3x^2-4x-2h-3hx}{x^2(x+h)^2}$

$=\frac{-3x^2-4x}{x^4}$

Then reduce,

$=\frac{-3x-4}{x^3}=-\frac{3x+4}{x^3}$

**Stoot** · June 6th 2012, 06:58 PM

Thanks for your help Reckoner. I was going wrong when multiplying the fraction by 1/h, silly really. Yeah, as you say: It's all about building solid foundations. It's all clear now!

Cheers,

Stoot

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