# Increasing function, partial derivatives

• Jul 13th 2009, 01:34 PM
arbolis
Increasing function, partial derivatives
I must say the veracity of the following statement : If $\displaystyle f:\mathbb{R}^2 \to \mathbb{R}$ differentiable such that $\displaystyle f_x(x,y)>0$ and $\displaystyle f_y(x,y)>0 \forall (x,y) \in \mathbb{R}^2$, then $\displaystyle g(t)=f(t,t^3)$ is an increasing function.
My attempt : I don't know how to start. By intuition it's false, so that the exercise show me a difference between one variable functions and several variables functions. I've not found any counter example yet. So it might be true, but I don't know how to prove it.
• Jul 13th 2009, 02:44 PM
Jester
Quote:

Originally Posted by arbolis
I must say the veracity of the following statement : If $\displaystyle f:\mathbb{R}^2 \to \mathbb{R}$ differentiable such that $\displaystyle f_x(x,y)>0$ and $\displaystyle f_y(x,y)>0 \forall (x,y) \in \mathbb{R}^2$, then $\displaystyle g(t)=f(t,t^3)$ is an increasing function.
My attempt : I don't know how to start. By intuition it's false, so that the exercise show me a difference between one variable functions and several variables functions. I've not found any counter example yet. So it might be true, but I don't know how to prove it.

Try showing $\displaystyle g'(t) > 0$.
• Jul 13th 2009, 07:23 PM
arbolis
Quote:

Originally Posted by Danny
Try showing $\displaystyle g'(t) > 0$.

My attempt : We know that $\displaystyle \frac{\partial f(x,y)}{\partial x}>0$ in particular for $\displaystyle x=t$ and $\displaystyle y=t^3$ since it's worth $\displaystyle \forall (x,y)\in \mathbb{R}^2$.
With the same spirit, one has $\displaystyle \frac{\partial f(x,y)}{\partial y}>0$ for $\displaystyle x=t$ and $\displaystyle y=t^3$.
If I'm not wrong, $\displaystyle \frac{df}{dt}(x,y)=$$\displaystyle \frac{\partial f(x,y)}{\partial x}+\frac{\partial f(x,y)}{\partial y}>0$. In particular $\displaystyle \frac{\partial f(t,t^3)}{\partial x} \cdot \frac{\partial x}{\partial t}+\frac{\partial f(t,t^3)}{\partial y}\cdot \frac{\partial y}{\partial t}>0 \Leftrightarrow \frac{df}{dt}(t,t^3)=g'(t)>0$, hence $\displaystyle g(t,t^3)$ is increasing.

I'm not confident at all. I doubt I had to use the chain rule! What do you say?