Increasing function, partial derivatives

**arbolis** · July 13th 2009, 01:34 PM

I must say the veracity of the following statement : If $f:\mathbb{R}^2 \to \mathbb{R}$ differentiable such that $f_x(x,y)>0$ and $f_y(x,y)>0 \forall (x,y) \in \mathbb{R}^2$ , then $g(t)=f(t,t^3)$ 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.
Any tip will be appreciated. (Not a full answer please

)

**Danny** · July 13th 2009, 02:44 PM

Originally Posted by arbolis

I must say the veracity of the following statement : If $f:\mathbb{R}^2 \to \mathbb{R}$ differentiable such that $f_x(x,y)>0$ and $f_y(x,y)>0 \forall (x,y) \in \mathbb{R}^2$ , then $g(t)=f(t,t^3)$ 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.
Any tip will be appreciated. (Not a full answer please

)

Try showing $g'(t) > 0$ .

**arbolis** · July 13th 2009, 07:23 PM

Originally Posted by Danny

Try showing $g'(t) > 0$ .

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

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

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