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Hi! Me again

Anyways. Yeah. I'm differenting from first principles and I'm having trouble with certain functions!

I'll give a simple example, and then hopefully, with your help, I can master these particular functions.

Differentiate, from first principles $f(x) = \frac{1}{x}$

I get:

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$

What's to be done here then? Would I use the index laws to get rid of the fractions on top, as such:

$f'(x) = \lim_{h \to 0} \frac{(x+h)^{-1}-x^{-1}}{h}$

Then multiply this by $\frac{(x+h)^{-1}+x^{-1}}{(x+h)^{-1}+x^{-1}}$

$f'(x) = \lim_{h \to 0} \frac{(x+h)^{-1}-x^{-1}}{h} \times \frac{(x+h)^{-1}+x^{-1}}{(x+h)^{-1}+x^{-1}}$

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

When I multiply this out past this point I get ridiculous and unhelpful fractions! Any ideas?!

Anyways. Yeah. I'm differenting from first principles and I'm having trouble with certain functions!

I'll give a simple example, and then hopefully, with your help, I can master these particular functions.

Differentiate, from first principles $f(x) = \frac{1}{x}$

I get:

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$

What's to be done here then? Would I use the index laws to get rid of the fractions on top, as such:

$f'(x) = \lim_{h \to 0} \frac{(x+h)^{-1}-x^{-1}}{h}$

Then multiply this by $\frac{(x+h)^{-1}+x^{-1}}{(x+h)^{-1}+x^{-1}}$

$f'(x) = \lim_{h \to 0} \frac{(x+h)^{-1}-x^{-1}}{h} \times \frac{(x+h)^{-1}+x^{-1}}{(x+h)^{-1}+x^{-1}}$

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

When I multiply this out past this point I get ridiculous and unhelpful fractions! Any ideas?!

Try writing $\displaystyle \frac{1}{x+h} - \frac{1}{x}$ as a single fraction over a common denominator.

Try shoving that on the top of the fraction you had at the third line from the bottom, before you multiplied it by some crazy fraction that = 1

Mush

$\frac{-h}{x(x+h)}$

$=\frac{-h}{x^{2}+hx}$

Then what

$=\frac{-h}{x^{2}+hx}$

Then what

What do you think? Put it back into the thing you were trying to work out.

Mush

$\frac{1}{x+h}-\frac{1}{x}=\frac{-h}{x(x+h)}$

Then what

Then what

Then $\lim_{h \rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h \rightarrow 0} \frac{\frac{-h}{x(x+h)}}{h}$, does it not?

Hi, ive got a problem with a question, can anyone help me out?

Differentiate from first principles:

$f(x) = \sqrt{x^2-1}$

ive tried:

$\lim_{h\rightarrow0} \frac{(x^2+h^2+2xh-1)^{1/2}-(x^2-1)^{1/2}}{h}$

but i'm now stuck!!

Differentiate from first principles:

$f(x) = \sqrt{x^2-1}$

ive tried:

$\lim_{h\rightarrow0} \frac{(x^2+h^2+2xh-1)^{1/2}-(x^2-1)^{1/2}}{h}$

but i'm now stuck!!

Try multiplying by $\frac{\sqrt{(x+h)^2-1}+\sqrt{x^2-1}}{\sqrt{(x+h)^2-1}+\sqrt{x^2-1}}$

cheers got it now

Original post by tonka121

Could someone help me with this one? From first principles, find the derivative of 1/(sqrt(x^2-1))

do you know the basic principles?

Original post by tonka121

Could someone help me with this one? From first principles, find the derivative of 1/(sqrt(x^2-1))

Please do not resurrect old threads - it is against the rules

Original post by tonka121

Sorry - I hadn't actually seen that it was so old!

Please start a fresh thread

Please post any work you have done

please quote in your replies because I am working also and I need to see notifications in order to respond

(I do not understand why resurrecting a thread is a problem.

Why not lock them then?)

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