What is the derivative of (1-tan(x)/(1+tan(x))) ?

The derivative of (1-tan(x)/(1+tan(x))) is -(2sec^2(x))/((1+tan(x))^2)

What is the first derivative of (1-tan(x)/(1+tan(x))) ?

The first derivative of (1-tan(x)/(1+tan(x))) is -(2sec^2(x))/((1+tan(x))^2)

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

\frac{10 x}{x ^{2} - 6}

f (x) = x^{2} ln (x)

\int_{0}^{3} x^{3} - 6 x d x

\int e^{- \frac{x}{50}} d x

y^{^{''}} + 20 y^{^{'}} + 200 y = 150

What is the derivative of (1-tan(x)/(1+tan(x))) ?
The derivative of (1-tan(x)/(1+tan(x))) is -(2sec^2(x))/((1+tan(x))^2)
What is the first derivative of (1-tan(x)/(1+tan(x))) ?
The first derivative of (1-tan(x)/(1+tan(x))) is -(2sec^2(x))/((1+tan(x))^2)