What is the derivative of (9x)/(8-tan(x)) ?

The derivative of (9x)/(8-tan(x)) is (9(8-tan(x)+xsec^2(x)))/((8-tan(x))^2)

What is the first derivative of (9x)/(8-tan(x)) ?

The first derivative of (9x)/(8-tan(x)) is (9(8-tan(x)+xsec^2(x)))/((8-tan(x))^2)

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derivative of

\frac{9 x}{8 - tan ( x )}

\frac{9 ( 8 - tan ( x ) + x sec ^{2} ( x ) )}{( 8 - tan ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 x}{8 - tan ( x )})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{x}{8 - tan ( x )})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d x}{d x} ( 8 - tan ( x ) ) - \frac{d}{d x} ( 8 - tan ( x ) ) x}{( 8 - tan ( x ) ) ^{2}}

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

\frac{d}{d x} (8 - tan (x)) = - sec^{2} (x)

= 9 \cdot \frac{1 \cdot ( 8 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 8 - tan ( x ) ) ^{2}}

Simplify

9 \cdot \frac{1 \cdot ( 8 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 8 - tan ( x ) ) ^{2}} : \frac{9 ( 8 - tan ( x ) + x sec ^{2} ( x ) )}{( 8 - tan ( x ) ) ^{2}}

= \frac{9 ( 8 - tan ( x ) + x sec ^{2} ( x ) )}{( 8 - tan ( x ) ) ^{2}}

\frac{\partial}{\partial v} (sin (u))

\frac{\partial}{\partial x} (\frac{( 5 x + 3 )}{y - 2})

\int_{3}^{5} (x - 3 + \frac{1}{x - 2}) d x

\frac{2 s + 1}{s ^{2} + 2 s + 1} \cdot \frac{1}{s}

x = \frac{3}{y}, x = \frac{y}{9}, x = 9 y

What is the derivative of (9x)/(8-tan(x)) ?
The derivative of (9x)/(8-tan(x)) is (9(8-tan(x)+xsec^2(x)))/((8-tan(x))^2)
What is the first derivative of (9x)/(8-tan(x)) ?
The first derivative of (9x)/(8-tan(x)) is (9(8-tan(x)+xsec^2(x)))/((8-tan(x))^2)