Question

what is the derivative of (sinx)^(tanx)

Answer 1

\[\ln y = \tan x~\ln(\sin(x))\]
might be simpler to play with

Answer 2

let\[y=(\sin x)^{\tan x}\Rightarrow \ln y= \tan x \ln \sin x \Rightarrow \frac{ y' }{ y }=\sec^2 x \ln \tan x+\tan x \frac{ \cos x }{ \sin x }\]\[=\sec^2 x \ln \sin x +1 \Rightarrow y' = \left( \sin x ^{\tan x} \right)\left( \sec^2 x \ln \sin x +1 \right)\]

Answer 3

where did the ln tanx come in from after the secx^2?

Answer 4

product rule is: (ab)' = a'b+ab'

Answer 5

looks to be a possible typo ...

Answer 6

do you know the equation to find the derivative of ln? isin't it something like 1/(1+x^2)?

Answer 7

ln is a function ...

ln(x) is a function of x

by the chain rule; ln(x) derives to x'/x

if this derivative is wrt.x then x' = dx/dx = 1

Answer 8

i think arctan(x) derives to 1/(1+x^2)

Answer 9

y = ln(x)

e^y = e^ln(x)

e^y = x

taking the derivative

e^y y' = x'

y' = x'/e^y , but e^y = x

y' = x'/x

Answer 10

so what would the overall correct answer be?

Answer 11

y = arctan(x)

tan(y) = x

sec^2(y) y' = x'

y' = x'/sec^2(y)

y' = x'/(1+tan^2(y)) ... tan y = x

y' = x'/(1+x^2)

Answer 12

work it out .... pg already did it for you with a typo or two :/

Answer 13

alright, thanks for the help!

Answer 14

just so you know:

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Answer 15

good to know, thank you