What is the derivative of (6x)/(1-tan(x)) ?

The derivative of (6x)/(1-tan(x)) is (6(1-tan(x)+xsec^2(x)))/((1-tan(x))^2)

What is the first derivative of (6x)/(1-tan(x)) ?

The first derivative of (6x)/(1-tan(x)) is (6(1-tan(x)+xsec^2(x)))/((1-tan(x))^2)

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derivative of

\frac{6 x}{1 - tan ( x )}

\frac{6 ( 1 - tan ( x ) + x sec ^{2} ( x ) )}{( 1 - tan ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{1 - tan ( x )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d x} (\frac{x}{1 - tan ( x )})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d x}{d x} ( 1 - tan ( x ) ) - \frac{d}{d x} ( 1 - tan ( x ) ) x}{( 1 - tan ( x ) ) ^{2}}

\frac{d x}{d x} = 1

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

= 6 \cdot \frac{1 \cdot ( 1 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 1 - tan ( x ) ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( 1 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 1 - tan ( x ) ) ^{2}} : \frac{6 ( 1 - tan ( x ) + x sec ^{2} ( x ) )}{( 1 - tan ( x ) ) ^{2}}

= \frac{6 ( 1 - tan ( x ) + x sec ^{2} ( x ) )}{( 1 - tan ( x ) ) ^{2}}

\int \frac{x - 3}{x + 1} d x

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int \frac{9}{x ( x + 3 )} d x

\int ln (y) d y

\int 3 x^{2} - 24 x + 36 d x

What is the derivative of (6x)/(1-tan(x)) ?
The derivative of (6x)/(1-tan(x)) is (6(1-tan(x)+xsec^2(x)))/((1-tan(x))^2)
What is the first derivative of (6x)/(1-tan(x)) ?
The first derivative of (6x)/(1-tan(x)) is (6(1-tan(x)+xsec^2(x)))/((1-tan(x))^2)