What is the derivative of (x^2+6/(sin^4(x))) ?

The derivative of (x^2+6/(sin^4(x))) is 2csc^5(x)(xsin(x)-2cos(x)(x^2+6))

What is the first derivative of (x^2+6/(sin^4(x))) ?

The first derivative of (x^2+6/(sin^4(x))) is 2csc^5(x)(xsin(x)-2cos(x)(x^2+6))

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\frac{d}{d x} (\frac{x ^{2} + 6}{sin ^{4} ( x )})

2 csc^{5} (x) (x sin (x) - 2 cos (x) (x^{2} + 6))

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 6}{sin ^{4} ( x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 6 ) sin ^{4} ( x ) - \frac{d}{d x} ( sin ^{4} ( x ) ) ( x ^{2} + 6 )}{( sin ^{4} ( x ) ) ^{2}}

\frac{d}{d x} (x^{2} + 6) = 2 x

\frac{d}{d x} (sin^{4} (x)) = 4 sin^{3} (x) cos (x)

= \frac{2 x sin ^{4} ( x ) - 4 sin ^{3} ( x ) cos ( x ) ( x ^{2} + 6 )}{( sin ^{4} ( x ) ) ^{2}}

Simplify

\frac{2 x sin ^{4} ( x ) - 4 sin ^{3} ( x ) cos ( x ) ( x ^{2} + 6 )}{( sin ^{4} ( x ) ) ^{2}} : 2 csc^{5} (x) (x sin (x) - 2 cos (x) (x^{2} + 6))

= 2 csc^{5} (x) (x sin (x) - 2 cos (x) (x^{2} + 6))

f (x) = - 9 e^{x^{- 9}}

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

x \to - 2 lim (3 x^{2} + 4 x - 2)

\frac{d y}{d x} = e^{4 x + 3 y}

What is the derivative of (x^2+6/(sin^4(x))) ?
The derivative of (x^2+6/(sin^4(x))) is 2csc^5(x)(xsin(x)-2cos(x)(x^2+6))
What is the first derivative of (x^2+6/(sin^4(x))) ?
The first derivative of (x^2+6/(sin^4(x))) is 2csc^5(x)(xsin(x)-2cos(x)(x^2+6))