What is the derivative of-1/(2sin^2(x)) ?

The derivative of-1/(2sin^2(x)) is csc^3(x)cos(x)

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

The first derivative of-1/(2sin^2(x)) is csc^3(x)cos(x)

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

csc^{3} (x) cos (x)

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{2}{( sin ( x ) ) ^{3}} \frac{d}{d x} (sin (x))

= - \frac{2}{( sin ( x ) ) ^{3}} \frac{d}{d x} (sin (x))

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

= - \frac{1}{2} (- \frac{2}{( sin ( x ) ) ^{3}} cos (x))

Simplify

- \frac{1}{2} (- \frac{2}{sin ^{3} ( x )} cos (x)) : csc^{3} (x) cos (x)

= csc^{3} (x) cos (x)

\int \frac{2 ^{x} \cdot 3 ^{x}}{9 ^{x} - 4 ^{x}} d x

\frac{d}{d x} (sin (x) + 2 cos (x))

x \to 3 lim (\frac{6 x}{x - 3})

\int \frac{1}{1 - ( - x )} d x

x d y - y d x = x \cdot tan (\frac{y}{x}) d x

What is the derivative of-1/(2sin^2(x)) ?
The derivative of-1/(2sin^2(x)) is csc^3(x)cos(x)
What is the first derivative of-1/(2sin^2(x)) ?
The first derivative of-1/(2sin^2(x)) is csc^3(x)cos(x)