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I have $$ f(x) = \frac 2x $$

and I have to differentiate from first principle. I have gotten to putting $ f(x) $ and $f(x+h) $ into the equation and I got $$ f'(x) = \lim_{h\to 0} \left\{ \frac{ -2}{x^2 + x} \right\}$$ However the correct answer doesn't have the $ + x $ on the bottom and i cannot work out how to get rid of it.

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I'd be curious to know how you got from $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ to the expression you show in your question. –  paw88789 Aug 29 at 10:36
    
for $ f(x+h) $ i got $ \frac 2(x+h) $ which after putting into the equation i ended up with $ \frac{ \frac 2(x+h) - \frac 2x }{h} $ after i made the denominators the same and done the subtraction i ended up with $ \frac { \frac{ -2h }{ x(x+h) } }{h} $ after removing the h i got the expression in the question –  Paul Aug 29 at 10:40
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That looks fine. And then you can cancel the topmost $h$ with the bottommost $h$ (the $h$ in the middle level stays in the expression). –  paw88789 Aug 29 at 10:42
    
Ah that would be the catch, i had expanded the middle bracket and took the h out with it so when i had $$ \lim_{h\to 0} $$ it did not take away the x. –  Paul Aug 29 at 10:48
    
Good. Glad you got that worked out! –  paw88789 Aug 29 at 10:56

2 Answers 2

up vote 10 down vote accepted

You should end up with

$$ \begin{align} f'(x) &=\lim_{h \to 0}\frac{f(x+h)-f(x)}h\\ &= \lim_{h \to 0}\frac{\frac2{x+h}-\frac2x}h\\ &=\lim_{h \to 0}\frac{2x-2(x+h)}{hx(x+h)}\\ &=\lim_{h \to 0}\frac{-2h}{hx(x+h)}\\ &=\lim_{h\to 0}\frac{-2}{x(x+h)} \end{align} $$

rather than what you have. Can you solve from here?

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

$$\equiv\lim_{h \to 0} \left[\frac{\frac{2}{x+h}-\frac{2}{x}}{h} \cdot \underbrace{\frac{x(x+h)}{x(x+h)}}_{1}\right]$$ $$\equiv\lim_{h \to 0} \left[\frac{2x-2(x+h)}{hx(x+h)}\right]$$

$$\require{cancel} \equiv\lim_{h \to 0}\left[\frac{-2\cancel{h}}{\cancel{h}x(x+h)}\right]$$

$$\equiv \lim_{h \to 0}\left[\frac{-2}{x^2+\cancelto{0}{xh}}\right]$$

$$ \equiv\Large \boxed{-\frac{2}{x^2}}$$

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