1/√x can be rewritten as ½x^-½. This can be integrated simply by the power rule for integration: ∫½x^-½dx = x^½ or √x.

]]>f(x) = f(g)/f(h), f'(x) = [ f'(g)f(h) - f(g)f'(h) ] / (f(h))²

In your case;

f(g) = 1, f(h) = √x, f'(g) = 0, f'(h) = 1/(2√x)

So, using the formula for differentiating fractions above, this all becomes;

[ 0(√x) - 1(1/(2√x))] / (√x)²;

This equals;

-1/(2x√x) = (-1/2)x^(-3/2)

Note that this type of discussion belongs in the "HELP ME!" section. Please post your math questions there in the future.

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