Asked by <3
Let f(x) = ln(((2x–5) / (7x+4))^(1/2)). f'(x) = ...
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Answered by
drwls
Let u = [(2x–5) / (7x+4))^(1/2)].
Use the "function of a function" or "chain rule", followed by the "quotent rule".
df/dx = f'(x) = d/du (ln u) * (du/dx)
= (1/u) * d/dx[(2x–5)/(7x+4))^(1/2)].
Finish by applying the quotient rule to find
d/dx[(2x–5)/(7x+4))^(1/2)].
Use the "function of a function" or "chain rule", followed by the "quotent rule".
df/dx = f'(x) = d/du (ln u) * (du/dx)
= (1/u) * d/dx[(2x–5)/(7x+4))^(1/2)].
Finish by applying the quotient rule to find
d/dx[(2x–5)/(7x+4))^(1/2)].
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