Question

if y = (1-cos x) (1 + tan x ). find dy/dx​

Answer 1

Answer:

$Y = (1 - cosx)(1 + tanx)$

$Y = UV{}^{1} + VU {}^{1}$

$\frac{dy}{dx } = (1 - cosx)\frac{d1 + tanx)}{dx} + (1 + tanx ) . \frac{d(1 - cosx )}{dx}$

$\frac{dy}{dx} =( 1 - cosx ).(0 + Sec {}^{2}x) + (1 + tanx ).(0 - ( - sinx )$

$\frac{dy}{dx} =( 1 - cosx ).sec{}^{2} x + (1 + tanx ).sinx$

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