已知\( y = \ln (x + 1) \)，则\( \frac{dy}{dx}\left| {_{x = 0}} \right. \)=______ 。
举一反三
- 函数\(z = {\left( {xy} \right)^x}\)的全微分为 A: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + x{\left( {xy} \right)^x}dy\) B: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) C: \(dz = {\left( {xy} \right)^x}\ln xydx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) D: \(dz = {\left( {xy} \right)^x}\left( {1 + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\)
- 下列方程中，不是全微分方程的为（ ）。 A: \(\left( {3{x^2} + 6x{y^2}} \right)dx + \left( {6{x^2}y + 4{y^2}} \right)dy = 0\) B: \({e^y}dx + \left( {x \cdot {e^y} - 2y} \right)dy = 0\) C: \(y\left( {x - 2y} \right)dx - {x^2}dy = 0\) D: \(\left( { { x^2} - y} \right)dx - xdy = 0\)
- 1. $\int \frac{1}{x(1+x)} dx =$ A: \[\ln{(x)}-\ln{\left( x+1\right) }+C\] B: \[\ln{(x)}+\ln{\left( x+1\right) }+C\] C: \[x-\ln{\left( x+1\right) }+C\] D: \[-\ln{(x)}+\ln{\left( x+1\right) }+C\]
- 已知\( y = \ln \left| x \right| \)，则\( y' \)为( ). A: \( {1 \over {\left| x \right|}} \) B: \( {1 \over x} \) C: \( - {1 \over x} \) D: \( x \)
- 已知\( y = \ln (2{\rm{ + }}6x) \)，则\( y'\left| {_{x = 0}} \right. \)为 .______