I am trying to find the second derivative of the function(adsbygoogle = window.adsbygoogle || []).push({});

[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.

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Multivariable chain rule question

Loading...

Similar Threads - Multivariable chain rule | Date |
---|---|

I Chain rule in a multi-variable function | May 7, 2016 |

Chain rule and multivariable calculus | Nov 3, 2015 |

Multivariable Calculus - chain rule on vectors | Sep 10, 2013 |

Multivariable Chain rule for higher order derivatives | May 9, 2013 |

Chain Rule and Multivariable Calculus Question | Jan 11, 2012 |

**Physics Forums - The Fusion of Science and Community**