anonymous
  • anonymous
Derivative of (cosx)^x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
logarithmic differentiation section huh? |dw:1333332330999:dw|
anonymous
  • anonymous
first take the log get \[\ln(\cos(x)^x)=x\ln(\cos(x))\] then take the derivitive using product and chain rule get \[\ln(\cos(x))+x\times \frac{-\sin(x)}{\cos(x)}=\ln(\cos(x))-x\tan(x)\] then multply by the original function to finish
anonymous
  • anonymous
d/dx(cos(x))^(x) remember e^(ln(x)) = x e^(ln(cos(x)))^(x) so we can just take the derivative of xln(cos(x)) to answer this now take the derivative using chain rule and product rule xln(cos(x)) = g(x) = ln(x) g'(x) = 1/x s(x) = cos(x) s'(x) = -sin(x) thus 1(ln(cos(x))) + (-sin(x)/cos(x))x e^(ln(cos(x)) + -xtan(x))

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anonymous
  • anonymous
Satalite I just dont see how your solution is correct as you are taking the ln of only one side of the function
Hero
  • Hero
\[\frac{d}{dx} \cos(x)^x = \cos^{x}(x)(\log(\cos(x))-x \tan(x))\]
Hero
  • Hero
You're welcome

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