What is the derivative of (x^2-6x/(1+cos(x))) ?

The derivative of (x^2-6x/(1+cos(x))) is ((2x-6)(1+cos(x))+sin(x)(x^2-6x))/((1+cos(x))^2)

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

The first derivative of (x^2-6x/(1+cos(x))) is ((2x-6)(1+cos(x))+sin(x)(x^2-6x))/((1+cos(x))^2)

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\frac{d}{d x} (\frac{x ^{2} - 6 x}{1 + cos ( x )})

\frac{( 2 x - 6 ) ( 1 + cos ( x ) ) + sin ( x ) ( x ^{2} - 6 x )}{( 1 + cos ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 6 x}{1 + cos ( x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 6 x ) ( 1 + cos ( x ) ) - \frac{d}{d x} ( 1 + cos ( x ) ) ( x ^{2} - 6 x )}{( 1 + cos ( x ) ) ^{2}}

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

\frac{d}{d x} (1 + cos (x)) = - sin (x)

= \frac{( 2 x - 6 ) ( 1 + cos ( x ) ) - ( - sin ( x ) ) ( x ^{2} - 6 x )}{( 1 + cos ( x ) ) ^{2}}

Simplify

= \frac{( 2 x - 6 ) ( 1 + cos ( x ) ) + sin ( x ) ( x ^{2} - 6 x )}{( 1 + cos ( x ) ) ^{2}}

\int \frac{1}{x ^{2} + 4 x + 25} d x

\int x 4 - x d x

\int_{0}^{1} 2 y^{2} d y

ln (e^{- x} + x e^{- x})

\frac{d}{d x} (\frac{5 x}{x - 3})

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