What is expand (y'(x)+(1-sec(x))y^2)tan(x) ?

The solution to expand (y'(x)+(1-sec(x))y^2)tan(x) is tan(x)y'(x)+tan(x)y^2-tan(x)sec(x)y^2

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expand (y'(x)+(1-sec(x))y^2)tan(x)

expand

(y^{'} (x) + (1 - sec (x)) y^{2}) tan (x)

tan (x) y' (x) + tan (x) y^{2} - tan (x) sec (x) y^{2}

Solution steps

(y^{'} (x) + (1 - sec (x)) y^{2}) tan (x)

= tan (x) (y^{^{'}} (x) + (1 - sec (x)) y^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = tan (x), b = y^{'} (x), c = (1 - sec (x)) y^{2}

= tan (x) y^{^{'}} (x) + tan (x) (1 - sec (x)) y^{2}

Expand

tan (x) y^{2} (1 - sec (x)) : tan (x) y^{2} - tan (x) sec (x) y^{2}

= tan (x) y^{'} (x) + tan (x) y^{2} - tan (x) sec (x) y^{2}

- \frac{2 ( - x + 1 )}{x ^{3}}

(e^{\frac{jω}{2}} - e^{- \frac{jω}{2}})^{2}

cos (x^{2} + 3 y - 1) (2 x + 3 \frac{d y}{d x})

\frac{5 x ^{3} + 14 x ^{2} + 23 x + 6}{( s + 3 ) \cdot ( s - 3 ) \cdot ( s + 1 ) ^{2}}

1 + \frac{1}{s ( s + 2.2 )}

What is expand (y'(x)+(1-sec(x))y^2)tan(x) ?
The solution to expand (y'(x)+(1-sec(x))y^2)tan(x) is tan(x)y'(x)+tan(x)y^2-tan(x)sec(x)y^2