If f(x)=e^squ root of x ln squ root of x, find F'(x)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If your function is f(x) = e^sqrt[x*ln(sqrt x)], then it's first derivative will be calculated using the chain rule and the product rule.

f'(x) = e^sqrt[x*ln(sqrt x)]*[x*ln(sqrt x)]'/2*sqrt[x*ln(sqrt x)]

f'(x) = e^sqrt[x*ln(sqrt x)]*[x'*ln(sqrt x) + x*[ln(sqrt x)]']'/2*sqrt[x*ln(sqrt x)]

f'(x) = e^sqrt[x*ln(sqrt x)]*[ln(sqrt x) + x/2x)]/2*sqrt[x*ln(sqrt x)]

f'(x) = e^sqrt[x*ln(sqrt x)]*[ln(sqrt x) + 1/2)]/2*sqrt[x*ln(sqrt x)]

The requested 1st derivative of the function is:

f'(x) = {e^sqrt[x*ln(sqrt x)]}*[ln(sqrt x) + 1/2)]/2*sqrt[x*ln(sqrt x)]