(xsqrt(4-x^2))^'

解答

(x 4 - x^{2})^{^{'}}

\frac{- 2 x ^{2} + 4}{4 - x ^{2}}

求解步骤

(x 4 - x^{2})^{'}

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = 4 - x^{2}

= x^{^{'}} 4 - x^{2} + (4 - x^{2})^{^{'}} x

x^{'} = 1

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

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

化简

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

= \frac{- 2 x ^{2} + 4}{4 - x ^{2}}

n = 1 \sum \infty \frac{1 ^{n}}{n ^{7}}

x \to 0 lim (\frac{x ^{2}}{∣ x ∣})

y^{^{'}} = t^{2} y^{2}

d er i v a t i v e f (x) = 2 x^{3} - 6 x^{2}

\int \frac{sin ^{4} ( x )}{4} d x