Question

dy/dx= y tanx-y² secx​

Answer 1

d/dx(uv) = u dv/dx + v du/dx

= [y sec^2 x + tanx dy/dx] - [y^2 (secx tanx) + 2(secx)y dy/dx]

= y sec^2x + tanx dy/dx - y^2 secx tanx - 2y secx dy/dx

= tan dy/dx - 2y secx dy/dx + y sec^2x - y^2 secx tanx

= dy/dx [tanx - 2y secx] + ysecx [secx - y tanx]

= dy/dx [1 - tanx + 2y secx]

= y secx [secx - y tanx]

= ysecx [secx - y tanx] /1 - tanx + 2y secx

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