derivative of (sqrt(5x^4+x^2)/2)

解答

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

\frac{10 x ^{2} + 1}{2 5 x ^{2} + 1}

求解步骤

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

将常数提出:

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

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

使用连锁律:

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

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

\frac{d}{d x} (5 x^{4} + x^{2}) = 20 x^{3} + 2 x

= \frac{1}{2} \cdot \frac{1}{2 5 x ^{4} + x ^{2}} (20 x^{3} + 2 x)

化简

\frac{1}{2} \cdot \frac{1}{2 5 x ^{4} + x ^{2}} (20 x^{3} + 2 x) : \frac{10 x ^{2} + 1}{2 5 x ^{2} + 1}

= \frac{10 x ^{2} + 1}{2 5 x ^{2} + 1}

d er i v a t i v e 2 sec^{2} (x) tan (x)

d er i v a t i v e cot (x^{2} + π x)

in v erse l a pl a ce \frac{s + 1}{( s ^{2} + 2 ) ( s + 1 )}

d er i v a t i v e 9 sin (2 x)

t an g e n t f (x) = \frac{x}{4 + x ^{2}}