What is the derivative of \sqrt[4]{x^5-x^3-2} ?

The derivative of \sqrt[4]{x^5-x^3-2} is (5x^4-3x^2)/(4(x^5-x^3-2)^{3/4)}

What is the first derivative of \sqrt[4]{x^5-x^3-2} ?

The first derivative of \sqrt[4]{x^5-x^3-2} is (5x^4-3x^2)/(4(x^5-x^3-2)^{3/4)}

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\frac{5 x ^{4} - 3 x ^{2}}{4 ( x ^{5} - x ^{3} - 2 ) ^{\frac{3}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( x ^{5} - x ^{3} - 2 ) ^{\frac{3}{4}}} \frac{d}{d x} (x^{5} - x^{3} - 2)

= \frac{1}{4 ( x ^{5} - x ^{3} - 2 ) ^{\frac{3}{4}}} \frac{d}{d x} (x^{5} - x^{3} - 2)

\frac{d}{d x} (x^{5} - x^{3} - 2) = 5 x^{4} - 3 x^{2}

= \frac{1}{4 ( x ^{5} - x ^{3} - 2 ) ^{\frac{3}{4}}} (5 x^{4} - 3 x^{2})

Simplify

\frac{1}{4 ( x ^{5} - x ^{3} - 2 ) ^{\frac{3}{4}}} (5 x^{4} - 3 x^{2}) : \frac{5 x ^{4} - 3 x ^{2}}{4 ( x ^{5} - x ^{3} - 2 ) ^{\frac{3}{4}}}

= \frac{5 x ^{4} - 3 x ^{2}}{4 ( x ^{5} - x ^{3} - 2 ) ^{\frac{3}{4}}}

\int 3 x e^{4 x} d x

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

s^{2} + t^{2}

\int e^{c o s (π x)} sin (π x) d x

n = 0 \sum \infty (n) sin (\frac{1}{n})

What is the derivative of \sqrt[4]{x^5-x^3-2} ?
The derivative of \sqrt[4]{x^5-x^3-2} is (5x^4-3x^2)/(4(x^5-x^3-2)^{3/4)}
What is the first derivative of \sqrt[4]{x^5-x^3-2} ?
The first derivative of \sqrt[4]{x^5-x^3-2} is (5x^4-3x^2)/(4(x^5-x^3-2)^{3/4)}