Question

differentiate : (sinx)tanx + (cosx)secx with respect to x

Answer 1

This must be the ans

Answer 2

$y = sinx \: tanx + cosx \: secx \\ \\ y = sinx \: tanx + 1 \\ \\ \frac{dy}{dx} = sinx {sec}^{2} x + tanx cosx + 0 \\ \\ \frac{dy}{dx} = sinx {sec}^{2} x + sinx \\ \\ \frac{dy}{dx} = sinx( {sec}^{2} x + 1)$