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Popular >

(1/(1-cos(x)))^'

Solution

(\frac{1}{1 - cos ( x )})^{^{'}}

Solution

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

Solution steps

(\frac{1}{1 - cos ( x )})^{'}

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= ((1 - cos (x))^{- 1})^{^{'}}

Apply the chain rule:

- \frac{1}{( 1 - cos ( x ) ) ^{2}} (1 - cos (x))^{'}

= - \frac{1}{( 1 - cos ( x ) ) ^{2}} (1 - cos (x))^{'}

(1 - cos (x))^{'} = sin (x)

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

Simplify

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

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

Popular Examples

(\frac{1}{x ^{2} + 1})^{^{'}}

(\partial)/(\partial x)(sin(2yx))

\frac{\partial}{\partial x} (sin (2 y x))

(dy)/(dx)=(2y)/(5y-2x)

\frac{d y}{d x} = \frac{2 y}{5 y - 2 x}

integral of 3tan(3x)

\int 3 tan (3 x) d x

integral of (2x^2-2x-1)/(x^3-x^2)

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

Frequently Asked Questions (FAQ)

What is the (1/(1-cos(x)))^' ?
The (1/(1-cos(x)))^' is -(sin(x))/((1-cos(x))^2)
What is the first (1/(1-cos(x)))^' ?
The first (1/(1-cos(x)))^' is -(sin(x))/((1-cos(x))^2)