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minisue1

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Hi, got this question, 'use the definition of a derivative to find the derivative of y= x^2 + 1/x'.
I figure I need to find limit as h approaches 0 [f(x+h)-f(x)]/h.
This leads me to [(x+h)^2+1/(x+h)-(x^2+1/x)]/h.

What do I need to do next? Help please
 
minisue1 said:
Hi, got this question, 'use the definition of a derivative to find the derivative of y= x^2 + 1/x'.
I figure I need to find limit as h approaches 0 [f(x+h)-f(x)]/h.
This leads me to [(x+h)^2+1/(x+h)-(x^2+1/x)]/h.

What do I need to do next? Help please
Click to expand...

\(\displaystyle \displaystyle \frac{\frac{1}{x+h} \ + (x+h)^2 \ - \ \frac{1}{x} - x^2}{h}\)

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

now continue....
 
ok, so after a super long effort of working out, i end up with
lim as h approaches 0=(2x^2+1)/x^2......is this right??
 
so i did it again, and got 2x+1/x^2.

f(x+h)=(x+h)^2+1/(x+h)
f(x+h)-f(x)=(x+h)^2+1/(x+h)-x^2-1/x
=2xh+h^2+h/(x^2+xh)
as 1/(x+h)-1/x=h/x(x+h)

so [f(x+h)-f(x)]/h=[2xh+h^2+h/(x^2+xh)]/h
=h[2x+h+1/(x^2+xh)]/h
=2x+h+1/(x^2+xh)

so lim as h approaches 0=2x+1/x^2.......:/
 
minisue1 said:
so i did it again, and got 2x+1/x^2.

f(x+h)=(x+h)^2+1/(x+h)
f(x+h)-f(x)=(x+h)^2+1/(x+h)-x^2-1/x
=2xh+h^2+h/(x^2+xh)
as 1/(x+h)-1/x=h/x(x+h) ====> CAREFUL! = -h/[x(x+h)] (Do you see why?)

so [f(x+h)-f(x)]/h=[2xh+h^2+h/(x^2+xh)]/h
=h[2x+h+1/(x^2+xh)]/h
=2x+h+1/(x^2+xh)

so lim as h approaches 0=2x+1/x^2.......:/
Click to expand...

.
 
Aaaah, yes...ooops.

So therefore the answer rather is 2x-1/x^2...

Thank you srmichael!
 
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