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September 18, 2014

September 18, 2014

Posted by **John** on Tuesday, October 19, 2010 at 8:30pm.

- calculus -
**MathMate**, Tuesday, October 19, 2010 at 9:34pmGiven f(x)=1/√x

f'(x)

=Lim h->0 (f(x+h)-f(x))/h

=Lim h->0 (1/√(x+h)-1/√x))/h

subtract with common denominator

=Lim h->0 ((√x-√(x+h)/(h(√x √(x+h)))

multiply by conjugate of numerator, √(x)+√(x+h)

=Lim h->0 (x-(x+h))/(h(√x √(x+h))*(√x+&radic(x+h)))

subtract and cancel h

=Lim h->0 -1/(√x √(x+h)*(√x+&radic(x+h)))

Take limit h->0

=-1/x^{3/2}

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