First we convert the square source to exponent notation.
You are watching: How do you find the derivative of a square root
$fracddx sqrtf(x)= fracddx f(x)^frac12 $
Then take the derivative and also apply the chain rule. That exponent is $-frac12$, because that some reason the markup language is do it hard to see the an unfavorable sign.
$= frac12 f(x)^frac-12f"(x)$
Converting back to notation v a square root symbol...
$= frac12frac1sqrtf(x) f"(x)$
Another possible way is logarithmic differentiation $$g(x)=sqrtf(x)$$ $$logig(g(x)ig)=frac 12 logig(f(x)ig)$$ $$fracg"(x)g(x)=frac 12fracf"(x)f(x)$$ $$g"(x)=frac 12fracf"(x)f(x)g(x)=frac 12fracf"(x)f(x)sqrtf(x)=frac 12fracf"(x)sqrtf(x)$$
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