Question

y = I - tanx/
1+ tanx
Differentiate with repect to x.​

Answer 1

Answer:

Please see the attachment

Answer 2

$\huge\mathfrak\red{Answer}$

$y = \frac{1 - tanx}{ 1+tanx } \\ y = \frac{tan45 - tanx}{1 + tan45.tanx} \\ as \: we \: know \: that \\ \tan( \alpha - \beta ) = \frac{ \tan( \alpha ) - \tan( \beta ) }{1 + \tan( \alpha ) . \tan( \beta ) } \\ \\ so \: similiarly \\ \tan(45 - x) = \frac{ \tan45 - tanx }{1 + \tan 45.tanx} \\ \\ so \: now \: we \: got \: \\ y = \tan(45 - x) \\ \frac{dy}{dx} = \frac{d(tan(45 - x))}{dx} \\ \frac{dy}{dx} = {sec}^2(45 - x) \frac{d(45 - x)}{dx} \\ \frac{dy}{dx} = (- 1){sec}^2(45 - x) \\ \\ |answer| = -{\sec}^2(45 - x)$