Pochodne cząstkowe pierwszego rzędu

[$liwa]

Prosiłbym o sprawdzenie czy dobrze rozumuje:
\(\displaystyle{ z=ln(x+ \sqrt{x ^{2} + y^{2} })}\)
\(\displaystyle{ \frac{\partial z}{\partial x} = \frac{1}{x+ \sqrt{x ^{2}+y ^{2} } }*(1+ \sqrt{2x})}\)
\(\displaystyle{ \frac{\partial z}{\partial y} = \frac{1}{x+ \sqrt{x ^{2}+y ^{2} } }*(1+ \sqrt{2y})}\)

-- 18 kwietnia 2009, 18:27 --

\(\displaystyle{ z=ln(sinh \frac{x}{2y}}\)
\(\displaystyle{ \frac{\partial z}{\partial x} = \frac{1}{sinh \frac{x}{2y} }*cosh \frac{x}{2y}* \frac{1}{2y}}\)
\(\displaystyle{ \frac{\partial z}{\partial y} = \frac{1}{sinh \frac{x}{2y} }*cosh \frac{x}{2y}*(- \frac{1}{2}x)}\)

[Gotta]

a)
\(\displaystyle{ \frac{\partial z}{\partial x}=\frac{1}{x+\sqrt{x^2+y^2}}(1+\frac{x}{\sqrt{x^2+y^2}})}\)
\(\displaystyle{ \frac{\partial z}{\partial y}=\frac{1}{x+\sqrt{x^2+y^2}}\frac{y}{\sqrt{x^2+y^2}}}\)

\(\displaystyle{ z=\ln \sinh\frac{x}{2y}}\)
\(\displaystyle{ \frac{\partial z}{\partial x}=\frac{1}{\sinh\frac{x}{2y}}\cosh\frac{x}{2y}\cdot\frac{1}{2y}}\)
\(\displaystyle{ \frac{\partial z}{\partial y}=\frac{1}{\sinh\frac{x}{2y}}\cosh\frac{x}{2y}\cdot -\frac{x}{2y^2}}\)

[$liwa]

\(\displaystyle{ z=(xy) ^{siny}}\)
\(\displaystyle{ \frac{\partial z}{\partial x}=siny*y}\)
\(\displaystyle{ \frac{\partial z}{\partial y}=cosy*x}\)

[Gotta]

\(\displaystyle{ z=(xy)^{\sin t}}\)
\(\displaystyle{ \frac{\partial z}{\partial x}=\sin y\cdot (xy)^{\sin y-1}\y}\)
\(\displaystyle{ z=(xy)^{\sin t}=e^{\sin y\ln xy}}\)
\(\displaystyle{ \frac{\partial z}{\partial y}=e^{\sin y\ln xy}(\cos y\ln xy+\sin y\cdot\frac{1}{y}}\)

[$liwa]

\(\displaystyle{ z=(lnx) ^{cos(x-y)}}\)
\(\displaystyle{ \frac{\partial z}{\partial x}=e ^{(cos(x-y)lnx}*(-sin(x-y)*1* \frac{1}{x} *1}\)
\(\displaystyle{ \frac{\partial z}{\partial y}=e ^{(cos(x-y)lnx}*(-sin(x-y)*(-1)* \frac{1}{x}}\)

[Gotta]

\(\displaystyle{ \frac{\partial }{\partial x}=e^{\cos (x-y)\ln (\ln x)\cdot(-\sin (x-y)\ln (\ln x)+\cos (x-y)\frac{1}{\ln x}\cdot\frac{1}{x})}}\)
\(\displaystyle{ \frac{\partial }{\partial x}=(\ln x)^{\cos (x-y)}\ln (\ln x)\cdot (-\sin (x-y))(-1)}\)