(\partial)/(\partial x)(ln(1+x^2+y^2-z^2))

解答

\frac{\partial}{\partial x} (ln (1 + x^{2} + y^{2} - z^{2}))

\frac{2 x}{1 + x ^{2} + y ^{2} - z ^{2}}

求解步骤

\frac{\partial}{\partial x} (ln (1 + x^{2} + y^{2} - z^{2}))

将

y, z

视为常数

使用连锁律:

\frac{1}{1 + x ^{2} + y ^{2} - z ^{2}} \frac{\partial}{\partial x} (1 + x^{2} + y^{2} - z^{2})

= \frac{1}{1 + x ^{2} + y ^{2} - z ^{2}} \frac{\partial}{\partial x} (1 + x^{2} + y^{2} - z^{2})

\frac{\partial}{\partial x} (1 + x^{2} + y^{2} - z^{2}) = 2 x

= \frac{1}{1 + x ^{2} + y ^{2} - z ^{2}} \cdot 2 x

化简

\frac{1}{1 + x ^{2} + y ^{2} - z ^{2}} \cdot 2 x : \frac{2 x}{1 + x ^{2} + y ^{2} - z ^{2}}

= \frac{2 x}{1 + x ^{2} + y ^{2} - z ^{2}}

\frac{d ^{3} y}{d x ^{3}} - cos (x) = 0

\int_{6}^{8} \frac{28}{( x - 6 ) ^{3}} d x

\int r^{3} d r

\int \frac{2 x ^{2} + 5 x + 18}{x ( x ^{2} + 9 )} d x

s l o p e in t erce pt (- 9, - 9), (- 6, 0)