3
$\begingroup$

I have a question:

Consider a function $g:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is differentiable. Find the derivative of the function: $G(x)=[ g\left ( x,x^{2},...,x^{n} \right )]^{2}$ where $x\in \mathbb{R}$.

Here, G is a function of one variable, so I tried to apply the chain rule to find its derivative as follows: G^{'}\left ( x \right )=2H^{'}\left ( x \right )H(x) where $H\left ( u_{1},u_{2},...,u_{n} \right )=g\left ( x,x^{2},...,x^{n} \right )$. Now to find the derivative of $H$, I did the following: $\frac{d}{dx}H=\frac{\partial H}{\partial u_{1}}\frac{\partial u_{1}}{\partial x}+...\frac{\partial H}{\partial u_{n}}\frac{\partial u_{n}}{\partial x}.$ Does what I did make sense?

  • 0
    Here is my answer based on the above comments: $G^{'}\left ( x \right )=2.g\left ( x,x^{2},...,x^{n} \right ).\left [ \frac{\partial g}{\partial u_{1}}.1+\frac{\partial g}{\partial u_{2}}.2x +...+nx^{n-1}.\frac{\partial g}{\partial u_{n}} \right ]$. where : $u_{i}=x^{i}$. I am not convinced to the presence of the variables $u_{i}=x^{i}$ in my final answer. Does anyone have a better idea on how to solve the problem?2012-01-29

1 Answers 1

1

It's almost correct. You have a function $u(\cdot):\quad {\mathbb R}\to{\mathbb R}^n\ ,\qquad x\mapsto u(x):=(x,x^2,\ldots, x^n)$ and a second function $g:\quad {\mathbb R}^n\to {\mathbb R}\ ,\qquad (u_1,\ldots, u_n)\mapsto g(u_1,\ldots, u_n)\ .$ The real-valued function $H:x\mapsto H(x)$ is defined as the composition of the two: $H(x)\ :=\ g\bigl(u(x)\bigr)\ .$ By the chain rule the derivative of $H$ is given by H'(x)={\partial g\over\partial u_1} u_1'(x) +\ldots+ {\partial g\over\partial u_n} u_n'(x) =\nabla g\bigl(u(x)\bigr)\cdot u'(x)\ , where the $\cdot$ denotes the scalar product.

  • 0
    Your "final answer" is correct. You can write $g_k$ instead of ${\partial g\over\partial u_k}$ if you want to avoid the appearance of the letter $u$ "out of nowhere".2012-01-30