derivative of sqrt(x^2+y^2-9)

解答

\frac{d}{d x} (x^{2} + y^{2} - 9)

\frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9}

求解步骤

\frac{d}{d x} (x^{2} + y^{2} - 9)

使用连锁律:

\frac{1}{2 x ^{2} + y ^{2} - 9} \frac{d}{d x} (x^{2} + y^{2} - 9)

= \frac{1}{2 x ^{2} + y ^{2} - 9} \frac{d}{d x} (x^{2} + y^{2} - 9)

\frac{d}{d x} (x^{2} + y^{2} - 9) = 2 x + 2 y \frac{d y}{d x}

= \frac{1}{2 x ^{2} + y ^{2} - 9} (2 x + 2 y \frac{d y}{d x})

化简

\frac{1}{2 x ^{2} + y ^{2} - 9} (2 x + 2 y \frac{d y}{d x}) : \frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9}

= \frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9}

\frac{d}{d x} (sin^{2} (x) - cos (2 x))

x \to 4 + lim (5)

\int sin (ln (14 x)) d x

d er i v a t i v e (\frac{x}{2 ( x ^{2} - 1 ) ^{2} - 1})^{2}

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