# Lost differentiating a function

1. Oct 2, 2009

### o_damhsa

1. The problem statement, all variables and given/known data The problem statement, all variables and given/known data[/b]
Hello. I'm trying to solve a non-linear problem and I have been working through the notes on this pdf to try and understand the method before I use it but I get stuck at one of the steps. The pdf is here:
http://www.nrbook.com/a/bookfpdf/f9-7.pdf
I cannot follow how the author went from equation 9.7.8 to 9.7.9

2. Relevant equations
The author has the following equation:

$$g (\lambda) \equiv f(xold + \lambda p)$$
He differentiates this function with respect to $$\lambda$$ and gets the following
$$g'(\lambda) = \nabla f \cdot p$$

3. The attempt at a solution
My understanding is that the author differentiated with respect to $$\lambda$$ so they got the $$\nabla f$$ and then differentiated what was inside the function to get the $$p$$ value. But I don't see how it became a dot product, or am I just misreading it?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 2, 2009

### Billy Bob

Welcome to PF, o_damhsa

The pdf doesn't open for me, but here's how it looks from what you've said.

Call $$xold=\langle x_1,x_2,x_3\rangle$$ and $$p=\langle p_1,p_2,p_3 \rangle$$

Then $$g(\lambda)=f(x_1+\lambda p_1,x_2+\lambda p_2,x_3+\lambda p_3)$$

Now by the Chain Rule,

$$\frac{dg}{d\lambda}= \frac{\partial f}{\partial x}\frac{dx}{d\lambda} +\frac{\partial f}{\partial y}\frac{dy}{d\lambda} +\frac{\partial f}{\partial z}\frac{dz}{d\lambda}$$

$$\frac{dg}{d\lambda}= \frac{\partial f}{\partial x} p_1 +\frac{\partial f}{\partial y} p_2 +\frac{\partial f}{\partial z} p_3$$

$$\frac{dg}{d\lambda}= \left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right\rangle\cdot\langle p_1,p_2,p_3 \rangle$$

$$\frac{dg}{d\lambda}= \nabla f\cdot p$$

3. Oct 5, 2009

### o_damhsa

Hello Billy Bob,

Thank you very much for your reply! It makes perfect sense now

o_damhsa