# Multivariable chain rule question

1. Jan 7, 2010

### willy0625

I am trying to find the second derivative of the function

$$C:[0,1]^{2} \rightarrow [0,1] ,\quad \mbox{defined by }C=C(u,v)$$

evaluated at

$$u=F(x)=1-\exp(-\lambda_{1} x),\quad \lambda_{1} \geq 0$$

and

$$v=G(x)=1-\exp(-\lambda_{2} x),\quad \lambda_{2} \geq 0$$

First I work out the first derivative which is

$$\dfrac{dC}{dx} = \dfrac{\partial C}{\partial u}\dfrac{du}{dx}+\dfrac{\partial C}{\partial v}\dfrac{dv}{dx}$$

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

$$\dfrac{du}{dx}$$

with respect to u(and v)??

I would appreciate any reply. Thank you guys.

Last edited: Jan 7, 2010
2. Jan 7, 2010

### Landau

$$\dfrac{dC}{dx} = \dfrac{\partial C}{\partial u}u'+\dfrac{\partial C}{\partial v}v'$$
$$\dfrac{d^2C}{dx^2} = \left(\frac{d}{dx}\dfrac{\partial C}{\partial u}\right)u'+\frac{\partial C}{\partial u}u''+\left(\frac{d}{dx}\dfrac{\partial C}{\partial v}\right)v'+\frac{\partial C}{\partial v}v''$$
$$\left(\frac{d}{dx}\dfrac{\partial C}{\partial u}\right)$$
$$\left(\frac{d}{dx}\dfrac{\partial C}{\partial v}\right)$$