chris215
  • chris215
find dy/dx of y=sec sqrtx
Mathematics
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jamiebookeater
  • jamiebookeater
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johnweldon1993
  • johnweldon1993
\[\large y = sec(\sqrt{x})\] Like that?
chris215
  • chris215
yeah
johnweldon1993
  • johnweldon1993
Okay sorry about that wait, was helping others as well :) So we just need to utilize the chain rule here \[\large \frac{dy}{dx} sec(x^{1/2})\] If we let u = x^(1/2) \[\large \frac{dy}{dx}sec(u) = \frac{dy}{dx} sec(u) \frac{dy}{dx}u\] So the derivative of \(\large sec(u) = sec(u)tan(u)\) and the derivative of u which is x^(1/2) = 1/2(x)^(-1/2) So altogether we have *remember to change u back to x^(1/2) \[\large \frac{dy}{dx} sec(\sqrt{x}) = \frac{sec(\sqrt{x})tan(\sqrt{x})}{2\sqrt{x}}\]

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chris215
  • chris215
thank you!!

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