- #1
willy0625
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I am trying to find the second derivative of the function
[tex] C:[0,1]^{2} \rightarrow [0,1] ,\quad \mbox{defined by }C=C(u,v) [/tex]
evaluated at
[tex]u=F(x)=1-\exp(-\lambda_{1} x),\quad \lambda_{1} \geq 0 [/tex]
and
[tex]v=G(x)=1-\exp(-\lambda_{2} x),\quad \lambda_{2} \geq 0[/tex]
First I work out the first derivative which is
[tex]\dfrac{dC}{dx} = \dfrac{\partial C}{\partial u}\dfrac{du}{dx}+\dfrac{\partial C}{\partial v}\dfrac{dv}{dx}[/tex]
Now, I have trouble working out the second derivative because it looks like I have to used the chain rule again and there is product rule which involves differentiating
[tex]\dfrac{du}{dx}[/tex]
with respect to u(and v)??I would appreciate any reply. Thank you guys.
[tex] C:[0,1]^{2} \rightarrow [0,1] ,\quad \mbox{defined by }C=C(u,v) [/tex]
evaluated at
[tex]u=F(x)=1-\exp(-\lambda_{1} x),\quad \lambda_{1} \geq 0 [/tex]
and
[tex]v=G(x)=1-\exp(-\lambda_{2} x),\quad \lambda_{2} \geq 0[/tex]
First I work out the first derivative which is
[tex]\dfrac{dC}{dx} = \dfrac{\partial C}{\partial u}\dfrac{du}{dx}+\dfrac{\partial C}{\partial v}\dfrac{dv}{dx}[/tex]
Now, I have trouble working out the second derivative because it looks like I have to used the chain rule again and there is product rule which involves differentiating
[tex]\dfrac{du}{dx}[/tex]
with respect to u(and v)??I would appreciate any reply. Thank you guys.
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