What is the derivative of f(x)=(5+csc(x))/(9-csc(x)) ?

The derivative of f(x)=(5+csc(x))/(9-csc(x)) is -(14cot(x)csc(x))/((9-csc(x))^2)

What is the first derivative of f(x)=(5+csc(x))/(9-csc(x)) ?

The first derivative of f(x)=(5+csc(x))/(9-csc(x)) is -(14cot(x)csc(x))/((9-csc(x))^2)

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derivative of

f (x) = \frac{5 + csc ( x )}{9 - csc ( x )}

- \frac{14 cot ( x ) csc ( x )}{( 9 - csc ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5 + csc ( x )}{9 - csc ( x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 5 + csc ( x ) ) ( 9 - csc ( x ) ) - \frac{d}{d x} ( 9 - csc ( x ) ) ( 5 + csc ( x ) )}{( 9 - csc ( x ) ) ^{2}}

\frac{d}{d x} (5 + csc (x)) = - cot (x) csc (x)

\frac{d}{d x} (9 - csc (x)) = cot (x) csc (x)

= \frac{( - cot ( x ) csc ( x ) ) ( 9 - csc ( x ) ) - cot ( x ) csc ( x ) ( 5 + csc ( x ) )}{( 9 - csc ( x ) ) ^{2}}

Simplify

\frac{( - cot ( x ) csc ( x ) ) ( 9 - csc ( x ) ) - cot ( x ) csc ( x ) ( 5 + csc ( x ) )}{( 9 - csc ( x ) ) ^{2}} : - \frac{14 cot ( x ) csc ( x )}{( 9 - csc ( x ) ) ^{2}}

= - \frac{14 cot ( x ) csc ( x )}{( 9 - csc ( x ) ) ^{2}}

g

\int \frac{1}{9 - 49 x ^{2}} d x

6 t sin (π t)

\int \frac{x}{x - 4} d x

What is the derivative of f(x)=(5+csc(x))/(9-csc(x)) ?
The derivative of f(x)=(5+csc(x))/(9-csc(x)) is -(14cot(x)csc(x))/((9-csc(x))^2)
What is the first derivative of f(x)=(5+csc(x))/(9-csc(x)) ?
The first derivative of f(x)=(5+csc(x))/(9-csc(x)) is -(14cot(x)csc(x))/((9-csc(x))^2)