What is the derivative of y=(2x)/(9-tan(x)) ?

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

What is the first derivative of y=(2x)/(9-tan(x)) ?

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

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derivative of

y = \frac{2 x}{9 - tan ( x )}

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

Solution steps

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

Take the constant out:

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

= 2 \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}}

= 2 \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)

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

Simplify

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

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

\frac{d}{d x} (- π^{2} sin (π x))

\int_{y}^{1} 3 x d x

f (4 k) = x^{2} - 2 x

\int \frac{17 - 17 x}{1 - x} d x

\frac{d}{d x} (ln (x^{2} + y^{2} + z^{2}))

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