Definition Of Derivative

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jarettbrock

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It would be very much appreciated if someone could show a step by step proccess of this one problem by using the definition of derivative: (f(x+h)-f(x))/h as h approaches zero. The problem is this: y=x^(1/2)-x


I am sorry that it is not typed in with a square root sign but I do not know how to do that. I know how to solve it other ways but for this specific problem we have been asked to use the definition of derivatives, so if you could show me how you did it, it would be great. Thank you
 
jarettbrock said:
It would be very much appreciated if someone could show a step by step proccess of this one problem by using the definition of derivative: (f(x+h)-f(x))/h as h approaches zero. The problem is this: y=x^(1/2)-x


I am sorry that it is not typed in with a square root sign but I do not know how to do that. I know how to solve it other ways but for this specific problem we have been asked to use the definition of derivatives, so if you could show me how you did it, it would be great. Thank you
Click to expand...

Can you find derivative of y = ?x, using definition of derivative?
 
If you can't i would be very happy because it was on my test today, but i have no clue, i only assumed you could do most basic functions using the definition of derivatives..

I just worked out y=x^(1/2) and i got the right answer so you can find the answer the definition of derivative, but i still cant figure out the other one.
 
jarettbrock said:
If you can't i would be very happy because it was on my test today, but i have no clue, i only assumed you could do most basic functions using the definition of derivatives..
Click to expand...

Yo u can and that's how it is done in the begining...

\(\displaystyle \lim_{h to 0} \left [ \frac{\sqrt{x+h} - \sqrt{x}}{h}\right ]\,\)

\(\displaystyle = \lim_{h to 0} \left [ \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \right ]\)

\(\displaystyle = \lim_{h to 0} \left [ \frac{x+h - x}{h \cdot (\sqrt{x+h} + \sqrt{x})} \right ]\)

Now continue.....
 
\(\displaystyle f(x) \ = \ x^{1/2}-x, \ f(x+h) \ = \ (x+h)^{1/2}-(x+h)\)

\(\displaystyle f \ ' \ (x) \ = \ \lim_{h\to0}\frac{f(x+h)-f(x)}{h} \ = \ \lim_{h\to0}\frac{(x+h)^{1/2}-(x+h)-(x^{1/2}-x)}{h}\)

\(\displaystyle = \ \lim_{h\to0}\frac{(x+h)^{1/2}-x-h-x^{1/2}+x}{h} \ = \ \lim_{h\to0}\frac{(x+h)^{1/2}-(x^{1/2}+h)}{h}\)

\(\displaystyle = \ \lim_{h\to0}\frac{(x+h)^{1/2}-(x^{1/2}+h)}{h}*\frac{(x+h)^{1/2}+(x^{1/2}+h)}{(x+h)^{1/2}+(x^{1/2}+h)}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{x+h-(x^{1/2}+h)^{2}}{h[(x+h)^{1/2}+(x^{1/2}+h)]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{x+h-(x+2hx^{1/2}+h^{2})}{h[(x+h)^{1/2}+(x^{1/2}+h)]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{x+h-x-2hx^{1/2}-h^{2}}{h[(x+h)^{1/2}+(x^{1/2}+h)]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{h-2hx^{1/2}-h^{2}}{h[(x+h)^{1/2}+(x^{1/2}+h)]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{1-2x^{1/2}-h}{(x+h)^{1/2}+(x^{1/2}+h)} \ = \ \frac{1-2x^{1/2}}{2x^{1/2}} \ = \ \frac{1}{2x^{1/2}}-1. \ QED\)
 
thank you sir! I didnt realize to combine the top part of the fraction and pull out a negative to make it a binomial so it would cancel out. I really your help
 
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