What is the derivative of \sqrt[3]{(x^2+1/(x^2-1)}) ?

The derivative of \sqrt[3]{(x^2+1/(x^2-1)}) is -(4x)/(3(x^2+1)^{2/3)(x^2-1)^{4/3}}

What is the first derivative of \sqrt[3]{(x^2+1/(x^2-1)}) ?

The first derivative of \sqrt[3]{(x^2+1/(x^2-1)}) is -(4x)/(3(x^2+1)^{2/3)(x^2-1)^{4/3}}

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- \frac{4 x}{3 ( x ^{2} + 1 ) ^{\frac{2}{3}} ( x ^{2} - 1 ) ^{\frac{4}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( \frac{x ^{2} + 1}{x ^{2} - 1} ) ^{\frac{2}{3}}} \frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 1})

= \frac{1}{3 ( \frac{x ^{2} + 1}{x ^{2} - 1} ) ^{\frac{2}{3}}} \frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 1})

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

= \frac{1}{3 ( \frac{x ^{2} + 1}{x ^{2} - 1} ) ^{\frac{2}{3}}} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}})

Simplify

\frac{1}{3 ( \frac{x ^{2} + 1}{x ^{2} - 1} ) ^{\frac{2}{3}}} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}}) : - \frac{4 x}{3 ( x ^{2} + 1 ) ^{\frac{2}{3}} ( x ^{2} - 1 ) ^{\frac{4}{3}}}

= - \frac{4 x}{3 ( x ^{2} + 1 ) ^{\frac{2}{3}} ( x ^{2} - 1 ) ^{\frac{4}{3}}}

\int_{0}^{\frac{π}{2}} 6 sec^{4} (\frac{t}{2}) d t

\frac{d}{d x} (7 - 3 x^{2})

\frac{\partial}{\partial y} (x^{2} y^{2} z^{2})

\int \frac{ln ( x )}{x ( ln ^{2} ( x ) + 4 )} d x

y^{^{''}} + 36 y = 12 sin (6 t)

What is the derivative of \sqrt[3]{(x^2+1/(x^2-1)}) ?
The derivative of \sqrt[3]{(x^2+1/(x^2-1)}) is -(4x)/(3(x^2+1)^{2/3)(x^2-1)^{4/3}}
What is the first derivative of \sqrt[3]{(x^2+1/(x^2-1)}) ?
The first derivative of \sqrt[3]{(x^2+1/(x^2-1)}) is -(4x)/(3(x^2+1)^{2/3)(x^2-1)^{4/3}}