derivative of \sqrt[4]{x^5-x^3-2}

解答

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

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

求解步骤

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

使用连锁律:

\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})

化简

\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)

d er i v a t i v e 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})