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Mathematics 20 Online
OpenStudy (anonymous):

What is the derivative of g(x) = 1/ sqrt (x) ?

OpenStudy (bahrom7893):

g(x) = x^(-1/2) g'(x) = (-1/2)*x^(-3/2)

OpenStudy (bahrom7893):

g'(x) = -1/(2*x^(3/2))

OpenStudy (anonymous):

can you show me how u get this answer..i got - 1 /x as the answer..

OpenStudy (bahrom7893):

1/sqrt(x) = x^(-1/2) right?

OpenStudy (anonymous):

yes but i left it as square root because sqrt x and sqrt x will be x

OpenStudy (bahrom7893):

why are you multiplying this by the square root?

OpenStudy (anonymous):

1/ sqrt (x+h) - 1/ sqrt (x) x 1/h = sqrt x - sqrt (x+h) / (sqrt (x+h) (sqrt x) and then multiply by the conjugate which is sqrt x + sqrt x+h

OpenStudy (anonymous):

so here all the square roots will be cancelled out

OpenStudy (bahrom7893):

oh wait u're solving this using a limit definition?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

then how u do it with limit?

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