# function --- derivative (help!)

#### gomes but instead of f(x)=x^-2, let f(x) = x^(1/2 )

So, I did
[ (x+h)^1/2 - (x)^1/2 ]/ h
But im not sure how to simplify it. What would be my next step? Thanks alot #### harish21 but instead of f(x)=x^-2, let f(x) = x^(1/2 )

So, I did
[ (x+h)^1/2 - (x)^1/2 ]/ h
But im not sure how to simplify it. What would be my next step? Thanks alot wait..

$$\displaystyle f(x) = x^{-2} \implies f(x) = \frac{1}{x^2}$$

now using the definition

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

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

simplify this and get your proof

• gomes

#### skeeter

MHF Helper but instead of f(x)=x^-2, let f(x) = x^(1/2 )
$$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$

$$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$$

$$\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$$

$$\displaystyle \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$$

finish it.

• gomes

#### gomes

Thanks everyone $$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$

$$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$$

$$\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$$

$$\displaystyle \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$$

finish it.
Thanks, I'm having trouble with sin(2x), i think my double angle formulae, i might be doing something wrong/missing something. Any help? Most appreciated! #### gomes

Sorry, one more, im stuck  #### AllanCuz

Sorry, one more, im stuck  I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Resources/sincosfirstprinciples.pdf

If you have any questions we can answer them here (Hi)

#### harish21

Thanks! i understand it, but what about for sin2x? $$\displaystyle \lim_{h \to 0} \frac{sin(2x+2h)-sin(2x)}{h}$$

$$\displaystyle = \lim_{h \to 0} \frac{2}{2} \times \frac{sin(2x+2h)-sin(2x)}{h}$$

$$\displaystyle = \lim_{h \to 0} \frac{2(sin(2x+2h)-sin(2x))}{2h}$$

$$\displaystyle = \lim_{h \to 0} \frac{2(sin(X+H)-sin(X))}{H}$$ where X=2x and H=2h

$$\displaystyle = 2cos(X)$$ ; which you proved before doing this.

$$\displaystyle = 2cos(2x)$$