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Math Help - Increasing function, partial derivatives

  1. #1
    MHF Contributor arbolis's Avatar
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    Increasing function, partial derivatives

    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)
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  2. #2
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    Quote Originally Posted by arbolis View Post
    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.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Danny View Post
    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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