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

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

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

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

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derivative of

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 1 + lim (\frac{∣ x - 1 ∣}{1 - x})

\int \frac{θ}{y ^{2}} e^{- \frac{θ}{y}} d y

\int \frac{5 x ^{2} - 5 x}{2 x ^{3} - 3 x ^{2} + 12} d x

\int (2 x^{5} + 2 x^{2} + 5) d x

x^{2} - 6 x + 9

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