What is the derivative of-ln^4(2sin^2(x^3-1)) ?

The derivative of-ln^4(2sin^2(x^3-1)) is -(48sin^5(x)ln^3(2sin^6(x)-1)cos(x))/(2sin^6(x)-1)

What is the first derivative of-ln^4(2sin^2(x^3-1)) ?

The first derivative of-ln^4(2sin^2(x^3-1)) is -(48sin^5(x)ln^3(2sin^6(x)-1)cos(x))/(2sin^6(x)-1)

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\frac{d}{d x} (- ln^{4} (2 sin^{2} (x)^{3} - 1))

- \frac{48 sin ^{5} ( x ) ln ^{3} ( 2 sin ^{6} ( x ) - 1 ) cos ( x )}{2 sin ^{6} ( x ) - 1}

Solution steps

\frac{d}{d x} (- ln^{4} (2 (sin^{2} (x))^{3} - 1))

Simplify

- ln^{4} (2 (sin^{2} (x))^{3} - 1) : - ln^{4} (2 sin^{6} (x) - 1)

= \frac{d}{d x} (- ln^{4} (2 sin^{6} (x) - 1))

Take the constant out:

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

= - \frac{d}{d x} (ln^{4} (2 sin^{6} (x) - 1))

Apply the chain rule:

4 (ln (2 sin^{6} (x) - 1))^{3} \frac{d}{d x} (ln (2 sin^{6} (x) - 1))

= 4 (ln (2 sin^{6} (x) - 1))^{3} \frac{d}{d x} (ln (2 sin^{6} (x) - 1))

\frac{d}{d x} (ln (2 sin^{6} (x) - 1)) = \frac{12 sin ^{5} ( x ) cos ( x )}{2 sin ^{6} ( x ) - 1}

= - 4 (ln (2 sin^{6} (x) - 1))^{3} \frac{12 sin ^{5} ( x ) cos ( x )}{2 sin ^{6} ( x ) - 1}

Simplify

- 4 ln^{3} (2 sin^{6} (x) - 1) \frac{12 sin ^{5} ( x ) cos ( x )}{2 sin ^{6} ( x ) - 1} : - \frac{48 sin ^{5} ( x ) ln ^{3} ( 2 sin ^{6} ( x ) - 1 ) cos ( x )}{2 sin ^{6} ( x ) - 1}

= - \frac{48 sin ^{5} ( x ) ln ^{3} ( 2 sin ^{6} ( x ) - 1 ) cos ( x )}{2 sin ^{6} ( x ) - 1}

x y^{^{'}} = \frac{1}{x y}

\int (\frac{x ^{3}}{1 + x ^{4}}) d x

\frac{\partial}{\partial x} (e^{4 x y} ln (7 y))

\int_{0}^{100} 37 cos (\frac{x}{80}) d x

What is the derivative of-ln^4(2sin^2(x^3-1)) ?
The derivative of-ln^4(2sin^2(x^3-1)) is -(48sin^5(x)ln^3(2sin^6(x)-1)cos(x))/(2sin^6(x)-1)
What is the first derivative of-ln^4(2sin^2(x^3-1)) ?
The first derivative of-ln^4(2sin^2(x^3-1)) is -(48sin^5(x)ln^3(2sin^6(x)-1)cos(x))/(2sin^6(x)-1)