How do you differentiate $y = e^{x^{2}} \cdot ln (tan x)$ $y=e^(x^2)*ln(tanx)$ ?

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Answer 1 · 2018-06-01T08:07:16

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

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

$f’(y) = u * dv + v * du$

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

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

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

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

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

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

How do you differentiate y=e^(x^2)*ln(tanx)y=ex2⋅ln(tanx)y=e^(x^2)*ln(tanx)?