What is the solution for -derivative of y=sec(x)+tan(x) ?

The solution for -derivative of y=sec(x)+tan(x) is y=-ln(tan(x)+sec(x))+ln(cos(x))+c_{1}

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Find the Derivative - d/dx y=sec(x)+tan(x)

- \frac{d}{d x} y = sec (x) + tan (x)

y = - ln (tan (x) + sec (x)) + ln (cos (x)) + c_{1}

Solution steps

- \frac{d y}{d x} = sec (x) + tan (x)

Isolate

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - sec (x) - tan (x) d x

\int - sec (x) - tan (x) d x = - ln (tan (x) + sec (x)) + ln (cos (x)) + c_{1}

y = - ln (tan (x) + sec (x)) + ln (cos (x)) + c_{1}

x \to \infty lim (\frac{x}{2})

sin (θ)

- \frac{d}{d x} (csc (x))

- \frac{d}{d x} (\frac{1}{( 1 - x )})

y = cos (x)

What is the solution for -derivative of y=sec(x)+tan(x) ?
The solution for -derivative of y=sec(x)+tan(x) is y=-ln(tan(x)+sec(x))+ln(cos(x))+c_{1}