How do you differentiate #y=e^(x^2)*ln(tanx)#?

Question

1751 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-01T08:07:16

#color(brown )(y’ = (e^x)^2 * ((1/(sin x* cos x)) + 2x * ln(tan x))#

#y = ((e^x)^2) * ln (tan x)#

#f’(y) = u * dv + v * du#

#u = ((e^x)^2), du = ((e^x)^2) * 2x = 2x * (e^x)^2#

#v = ln(tan x), dv = (1/tan x) * sec^2 x#

#dv = (1/(sin x / cos x)) * (1/cos^2 x) = 1 / (sin x * cos x)#

#y’ = (e^x)^2 * (1 / (sin x * cos x)) + ln(tan x) * 2x * (e^x)^2#

#y’ = (e^x)^2 / (sin x * cos x) + 2x * (e^x)^2 * ln(tan x)#

#color(brown )(y’ = (e^x)^2 * ((1/(sin x* cos x)) + 2x * ln(tan x))#