Bounded Homework Derivative Answers

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SUMMARY

The discussion analyzes the boundedness of the derivatives of two functions defined on the domain [-1,1] x ℝ. For the function f(x,y) = 3exp(x-y²), the derivative with respect to y, ∂f/∂y = -6yexp(x-y²), is confirmed to be bounded due to the properties of the exponential function and the bounded nature of yexp(-y²). Conversely, for the function f(x,y) = 7exp(y²-x), the derivative ∂f/∂y = 14yexp(y²-x) is determined to be unbounded, as it grows without limit as y increases.

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Homework Statement



Let [tex]f:[-1,1] \times \mathbb{R} \to\mathbb{R}[/tex] be a function.

If [itex]f[/itex] is defined by:

(i) [itex]f(x,y) = 3\exp(x-y^2)[/itex]

then is the derivative with respect to [itex]y[/itex] bounded?

If [itex]f[/itex] is defined by:

(ii) [itex]f(x,y) = 7\exp(y^2-x)[/itex]

then is the derivative with respect to [itex]y[/itex] bounded?

The Attempt at a Solution



For (i):

[itex]\frac{\partial f}{\partial y} = -6y\exp(x-y^2) = -6y\exp(-y^2)\exp(x)[/itex]

and since [itex]\exp(x) \leq e[/itex] in the specified domain and since [itex]y\exp(-y^2)[/itex] is a bounded function on [itex]\mathbb{R}[/itex], [itex]\frac{\partial f}{\partial y}[/itex] is bounded - right?

For (ii):

[itex]\frac{\partial f}{\partial y} = 14y\exp(y^2-x) = 14y\exp(y^2)\exp(-x)[/itex]

This is not bounded - right?
 
Last edited:
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That looks right.
 

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