- #1

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y = (cosx)^sinx

any help wud be appreciated, im unsure of how to start this question, thx in advance

- Thread starter bengalibabu
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- #1

- 18

- 0

y = (cosx)^sinx

any help wud be appreciated, im unsure of how to start this question, thx in advance

- #2

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Logarithmate and differentiate using the chain rule.

Daniel.

Daniel.

- #3

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- #4

Curious3141

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[tex]a = e^{\ln{a}}[/tex] where 'ln' is natural log.

- #5

Curious3141

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- #6

Pyrrhus

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Apply logarithm on both sides, and use the property of logs [itex] \ln a^b = b \ln a [/itex]

- #7

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y = (cosx)^sinx

logy = log(cosx^sinx)

logy=sinx(logcosx)

(dy/dx)(1/(yln10)) = sinx(1/cosx(ln10) + log(cosx^cosx)

(dy/dx)=[(tanx/ln10) + (log(cosx^cosx))](yln10)

jus let me know if im on the right track. thanks for your help guys.

- #8

Pyrrhus

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[tex] \ln y = \sin x \ln (\cos x) [/tex]

[tex] \frac{dy}{dx} \frac{1}{y} = \sin x \frac{-\sin x}{\cos x} + \cos x \\ln (\cos x) [/tex]

[tex] \frac{dy}{dx} = (\cos x)^{\sin x}(\sin x \frac{-\sin x}{\cos x} + \cos x \ln (\cos x)) [/tex]

- #9

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isnt the derivative of log(base)x = 1/xln(base) ????

- #10

Pyrrhus

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Yes, you should have used natural logarithm to avoid that, but it's ok.

- #11

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how could u have used natural logs to avoid that?

- #12

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o nevermind stupid question lol. i should have converted log to ln then found the deriative.

- #13

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hah - "logarithmate"? I would prefer "logarithmize"

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