How do you use the definition of the derivative to find $f'(x)$ if $f(x)=1/sqrtx$ ?

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Answer 1 · 2016-09-28T03:18:51

Start with $lim/(h→0){f(x+h)-f(x)}/h$ multiply numerator and denominator by $f(x+h) + f(x)$

$lim/(h→0){(f(x+h))^2-(f(x))^2}/(h(f(x) + f(x+h)))$

We observe that $(f(x))^2 = 1/x$ and $(f(x+h))^2 = 1/(x+h)$ and we do the substitution:

$lim/(h→0){1/(x+h) - 1/x}/(h(1/sqrt(x) + 1/(sqrt(x+h)))$

Make a common denominator for $1/(x+h) - 1/x = (x)/((x(x +h))) - (x+h)/((x(x +h))) = -h/((x(x +h)))$

Substitute back into the limit:

$lim/(h→0){-h/((x(x +h)))}/(h(1/sqrt(x) + 1/(sqrt(x+h)))$

Please observe that $-h/h = -1$

$lim/(h→0){-1/((x(x +h)))}/((1/sqrt(x) + 1/(sqrt(x+h)))$

Now it is safe to let h go to zero:

${-1/((x(x)))}/((1/sqrt(x) + 1/(sqrt(x)))) = (-1/x^2)/(2/sqrt(x)) = -1/(2x^(3/2))$

How do you use the definition of the derivative to find f'(x)f'(x) if f(x)=1/sqrtxf(x)=1/sqrtx?