What is the derivative of (6x/(9-tan(x))) ?

The derivative of (6x/(9-tan(x))) is (6(9-tan(x)+xsec^2(x)))/((9-tan(x))^2)

What is the first derivative of (6x/(9-tan(x))) ?

The first derivative of (6x/(9-tan(x))) is (6(9-tan(x)+xsec^2(x)))/((9-tan(x))^2)

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\frac{d}{d x} (\frac{6 x}{9 - tan ( x )})

\frac{6 ( 9 - tan ( x ) + x sec ^{2} ( x ) )}{( 9 - tan ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{9 - tan ( x )})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{9 - 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} ( 9 - tan ( x ) ) - \frac{d}{d x} ( 9 - tan ( x ) ) x}{( 9 - tan ( x ) ) ^{2}}

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

\frac{d}{d x} (9 - tan (x)) = - sec^{2} (x)

= 6 \cdot \frac{1 \cdot ( 9 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 9 - tan ( x ) ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( 9 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 9 - tan ( x ) ) ^{2}} : \frac{6 ( 9 - tan ( x ) + x sec ^{2} ( x ) )}{( 9 - tan ( x ) ) ^{2}}

= \frac{6 ( 9 - tan ( x ) + x sec ^{2} ( x ) )}{( 9 - tan ( x ) ) ^{2}}

\frac{d}{d x} (e^{t a n (π x)})

\int \frac{40}{40 + e ^{x}} d x

f (x) = 2 e^{x - 1}

\int \frac{9}{x + 1} d x

\frac{d}{d x} (3 (- csc^{2} (x) + sec^{2} (x)))

What is the derivative of (6x/(9-tan(x))) ?
The derivative of (6x/(9-tan(x))) is (6(9-tan(x)+xsec^2(x)))/((9-tan(x))^2)
What is the first derivative of (6x/(9-tan(x))) ?
The first derivative of (6x/(9-tan(x))) is (6(9-tan(x)+xsec^2(x)))/((9-tan(x))^2)