What is derivative of arctan(sec(x+tan(x))) ?

The answer to derivative of arctan(sec(x+tan(x))) is -1/(((sec(x)+tan(x))^2+1)(-1+sin(x)))

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derivative of arctan(sec(x+tan(x)))

\frac{d}{d x} (arctan (sec (x) + tan (x)))

- \frac{1}{( ( sec ( x ) + tan ( x ) ) ^{2} + 1 ) ( - 1 + sin ( x ) )}

Solution steps

\frac{d}{d x} (arctan (sec (x) + tan (x)))

Apply the chain rule:

\frac{1}{( sec ( x ) + tan ( x ) ) ^{2} + 1} \frac{d}{d x} (sec (x) + tan (x))

= \frac{1}{( sec ( x ) + tan ( x ) ) ^{2} + 1} \frac{d}{d x} (sec (x) + tan (x))

\frac{d}{d x} (sec (x) + tan (x)) = - \frac{1}{- 1 + sin ( x )}

= \frac{1}{( sec ( x ) + tan ( x ) ) ^{2} + 1} (- \frac{1}{- 1 + sin ( x )})

Simplify

\frac{1}{( sec ( x ) + tan ( x ) ) ^{2} + 1} (- \frac{1}{- 1 + sin ( x )}) : - \frac{1}{( ( sec ( x ) + tan ( x ) ) ^{2} + 1 ) ( - 1 + sin ( x ) )}

= - \frac{1}{( ( sec ( x ) + tan ( x ) ) ^{2} + 1 ) ( - 1 + sin ( x ) )}

y = \frac{sin ( x )}{x ^{2}}

\frac{d}{d x} (\frac{5}{x + 1})

y^{^{'}} x y = x^{2} + 2 y^{2}

\frac{d}{d x} ((\frac{1 - 2 x}{1 + 3 x})^{4})

\int_{- 9}^{9} 1 + x^{2} d x

What is derivative of arctan(sec(x+tan(x))) ?
The answer to derivative of arctan(sec(x+tan(x))) is -1/(((sec(x)+tan(x))^2+1)(-1+sin(x)))