What is the derivative of ((4-x))/(sin^2(x)) ?

The derivative of ((4-x))/(sin^2(x)) is (-sin^2(x)-sin(2x)(4-x))/(sin^4(x))

What is the first derivative of ((4-x))/(sin^2(x)) ?

The first derivative of ((4-x))/(sin^2(x)) is (-sin^2(x)-sin(2x)(4-x))/(sin^4(x))

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derivative of

\frac{( 4 - x )}{sin ^{2} ( x )}

\frac{- sin ^{2} ( x ) - sin ( 2 x ) ( 4 - x )}{sin ^{4} ( x )}

Solution steps

\frac{d}{d x} (\frac{4 - x}{sin ^{2} ( x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 - x ) sin ^{2} ( x ) - \frac{d}{d x} ( sin ^{2} ( x ) ) ( 4 - x )}{( sin ^{2} ( x ) ) ^{2}}

\frac{d}{d x} (4 - x) = - 1

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

= \frac{( - 1 ) sin ^{2} ( x ) - sin ( 2 x ) ( 4 - x )}{( sin ^{2} ( x ) ) ^{2}}

Simplify

\frac{( - 1 ) sin ^{2} ( x ) - sin ( 2 x ) ( 4 - x )}{( sin ^{2} ( x ) ) ^{2}} : \frac{- sin ^{2} ( x ) - sin ( 2 x ) ( 4 - x )}{sin ^{4} ( x )}

= \frac{- sin ^{2} ( x ) - sin ( 2 x ) ( 4 - x )}{sin ^{4} ( x )}

y = ((e^{2} x) + 1)^{3}

e^{2} + x

∣ x + 2 ∣ - 2

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

\frac{d}{df} (f (1) + f (- 2))

What is the derivative of ((4-x))/(sin^2(x)) ?
The derivative of ((4-x))/(sin^2(x)) is (-sin^2(x)-sin(2x)(4-x))/(sin^4(x))
What is the first derivative of ((4-x))/(sin^2(x)) ?
The first derivative of ((4-x))/(sin^2(x)) is (-sin^2(x)-sin(2x)(4-x))/(sin^4(x))