What is the derivative of f(x)=(6cos(x))/(7-9cos(x)) ?

The derivative of f(x)=(6cos(x))/(7-9cos(x)) is -(42sin(x))/((7-9cos(x))^2)

What is the first derivative of f(x)=(6cos(x))/(7-9cos(x)) ?

The first derivative of f(x)=(6cos(x))/(7-9cos(x)) is -(42sin(x))/((7-9cos(x))^2)

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derivative of

f (x) = \frac{6 cos ( x )}{7 - 9 cos ( x )}

- \frac{42 sin ( x )}{( 7 - 9 cos ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 cos ( x )}{7 - 9 cos ( x )})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{cos ( x )}{7 - 9 cos ( x )})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d}{d x} ( cos ( x ) ) ( 7 - 9 cos ( x ) ) - \frac{d}{d x} ( 7 - 9 cos ( x ) ) cos ( x )}{( 7 - 9 cos ( x ) ) ^{2}}

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

\frac{d}{d x} (7 - 9 cos (x)) = 9 sin (x)

= 6 \cdot \frac{( - sin ( x ) ) ( 7 - 9 cos ( x ) ) - 9 sin ( x ) cos ( x )}{( 7 - 9 cos ( x ) ) ^{2}}

Simplify

6 \cdot \frac{( - sin ( x ) ) ( 7 - 9 cos ( x ) ) - 9 sin ( x ) cos ( x )}{( 7 - 9 cos ( x ) ) ^{2}} : - \frac{42 sin ( x )}{( 7 - 9 cos ( x ) ) ^{2}}

= - \frac{42 sin ( x )}{( 7 - 9 cos ( x ) ) ^{2}}

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

cos (2 x^{2})

\int_{0}^{2} \frac{3 ln ( x )}{x} d x

\frac{d}{d x} (\frac{1}{4} e^{x})

\frac{d}{d x} (ln (\frac{3 + x}{3 - x}))

What is the derivative of f(x)=(6cos(x))/(7-9cos(x)) ?
The derivative of f(x)=(6cos(x))/(7-9cos(x)) is -(42sin(x))/((7-9cos(x))^2)
What is the first derivative of f(x)=(6cos(x))/(7-9cos(x)) ?
The first derivative of f(x)=(6cos(x))/(7-9cos(x)) is -(42sin(x))/((7-9cos(x))^2)