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Calculus1 26 Online
OpenStudy (anonymous):

How do you find the derivative of (sin(x))^1/2? Would you use the power or chain rule?

OpenStudy (ranga):

power rule first and then chain rule.

OpenStudy (ranga):

Derivative of [ f(x) ]^n is: n * [ f(x) ]^(n-1) * f'(x)

OpenStudy (anonymous):

try chain rule may b easier for you or which one do you find complex to use?

OpenStudy (anonymous):

\[(\sin(x))^{1/2}\] First the power rule. \[\frac{ 1}{ 2 }(\sin(x))^{\frac{ 1 }{ 2 }-1}\] Then the chain rule to take the derivative of the inside. \[\frac{ 1}{ 2 }(\sin(x))^{-\frac{ 1 }{ 2 }}*\cos(x)\] Last step is to simplify \[\frac{ \cos(x) }{ 2\sqrt{\sin(x)} }\]

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