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I posted these a while back, I just wanted to be certain I did it right before I hand it in:
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Question 1
Suppose that h : \mathbb{R} \to \mathbb{R} is continuous. Calculate f', if f : \mathbb{R}^2 \to \mathbb{R} is the function:
f(x, y) = \int _{\sin (xy)} ^{\cos (xy)} h(t)dt
Question 2
Let
f(x,y)=\left\{\begin{array}{cc}\frac{x^4 + y^4}{x^2 + y^2},&\mbox{ if }<br /> (x,y)\neq (0,0)\\0, & \mbox{ if } (x,y) = (0,0)\end{array}\right
Show that f is differentiable at (0,0)
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Question 1
f' = \left [D_1f(x, y) \ \ \ \ D_2f(x, y)\right ]
f' = \left [\frac{\partial}{\partial x}\int_{\sin (xy)} ^{\cos (xy)} h(t)dt \ \ \ \ \frac{\partial}{\partial y}\int_{\sin (xy)} ^{\cos (xy)} h(t)dt\right ]
Let H be the antiderivative of h (which exists because h is continuous), then:
f' = \left [\frac{\partial}{\partial x}\left ( H(\cos xy) - H(\sin xy) \right ) \ \ \ \ \frac{\partial}{\partial y}\left ( H(\cos xy) - H(\sin xy)\right ) \right ]
f' = -\left [y(\sin (xy)h(\cos (xy)) + \cos (xy)h(\sin (xy))) \ \ \ \ x(\sin (xy)h(\cos (xy)) + \cos (xy)h(\sin (xy)))\right ]
Question 2
Proposition: f is differentiable at (0, 0), and the derivative at that point is the zero transformation.
Proof:
It suffices to show that:
L = 0
Where h = (h_1, h_2) \in \mathbb{R}^2, and:
L = \lim _{h \rightarrow 0} \frac{|f(h)|}{|h|}
If h_1 = 0 and h_2 \neq 0, then:
L = \lim _{h \rightarrow 0} \frac{h_2^4}{|h_2^3|} = \lim _{h \rightarrow 0} |h_2| = 0
Clearly, if h_1 \neq 0 and h_2 = 0, then we also have that L = 0. Now, if neither component of h is zero, then:
L = \lim _{h \rightarrow 0} \frac{|h_1^4 + h_2^4|}{(h_1^2 + h_2^2)^{3/2}}
L = \lim _{h \rightarrow 0} \frac{|(h_1^2 + h_2^2)^2 - 2h_1^2h_2^2|}{(h_1^2 + h_2^2)^{3/2}}
L = \lim _{h \rightarrow 0} \left ||h| - \frac{2h_1^2h_2^2}{(h_1^2 + h_2^2)^{3/2}} \right |
L = 2\lim _{h \rightarrow 0} \frac{h_1^2h_2^2}{(h_1^2 + h_2^2)^{3/2}}
L = 2\lim _{h \rightarrow 0} |h| \left (\frac{h_1h_2}{h_1^2 + h_2^2} \right )^2
Now, consider the function g(z) = z + z^{-1} for positive z \in \mathbb{R}. Simple analysis shows that g reaches a minimum at 2, so:
z + \frac{1}{z} \geq 2
Now, let |h_1||h_2|^{-1} = z:
\frac{|h_1|}{|h_2|} + \frac{|h_2|}{|h_1|} \geq 2
h_1^2 + h_2^2 \geq 2|h_1||h_2|
\frac{1}{2} \geq \frac{|h_1||h_2|}{h_1^2 + h_2^2}
\frac{1}{4} \geq \left ( \frac{h_1h_2}{h_1^2 + h_2^2} \right )^2
2|h|\frac{1}{4} \geq 2|h|\left ( \frac{h_1h_2}{h_1^2 + h_2^2} \right )^2
So:
L \leq \frac{1}{2}\lim _{h \rightarrow 0} |h| = 0
Clearly, |f(h)| and |h| are non-negative, so:
0 \leq \lim _{h \rightarrow 0} \frac{|f(h)|}{|h|} = L \leq 0[/itex]<br /> <br /> This proves that f is indeed differentiable at (0, 0), and 0 (the zero transformation) is its derivative there. Q.E.D.
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Question 1
Suppose that h : \mathbb{R} \to \mathbb{R} is continuous. Calculate f', if f : \mathbb{R}^2 \to \mathbb{R} is the function:
f(x, y) = \int _{\sin (xy)} ^{\cos (xy)} h(t)dt
Question 2
Let
f(x,y)=\left\{\begin{array}{cc}\frac{x^4 + y^4}{x^2 + y^2},&\mbox{ if }<br /> (x,y)\neq (0,0)\\0, & \mbox{ if } (x,y) = (0,0)\end{array}\right
Show that f is differentiable at (0,0)
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Question 1
f' = \left [D_1f(x, y) \ \ \ \ D_2f(x, y)\right ]
f' = \left [\frac{\partial}{\partial x}\int_{\sin (xy)} ^{\cos (xy)} h(t)dt \ \ \ \ \frac{\partial}{\partial y}\int_{\sin (xy)} ^{\cos (xy)} h(t)dt\right ]
Let H be the antiderivative of h (which exists because h is continuous), then:
f' = \left [\frac{\partial}{\partial x}\left ( H(\cos xy) - H(\sin xy) \right ) \ \ \ \ \frac{\partial}{\partial y}\left ( H(\cos xy) - H(\sin xy)\right ) \right ]
f' = -\left [y(\sin (xy)h(\cos (xy)) + \cos (xy)h(\sin (xy))) \ \ \ \ x(\sin (xy)h(\cos (xy)) + \cos (xy)h(\sin (xy)))\right ]
Question 2
Proposition: f is differentiable at (0, 0), and the derivative at that point is the zero transformation.
Proof:
It suffices to show that:
L = 0
Where h = (h_1, h_2) \in \mathbb{R}^2, and:
L = \lim _{h \rightarrow 0} \frac{|f(h)|}{|h|}
If h_1 = 0 and h_2 \neq 0, then:
L = \lim _{h \rightarrow 0} \frac{h_2^4}{|h_2^3|} = \lim _{h \rightarrow 0} |h_2| = 0
Clearly, if h_1 \neq 0 and h_2 = 0, then we also have that L = 0. Now, if neither component of h is zero, then:
L = \lim _{h \rightarrow 0} \frac{|h_1^4 + h_2^4|}{(h_1^2 + h_2^2)^{3/2}}
L = \lim _{h \rightarrow 0} \frac{|(h_1^2 + h_2^2)^2 - 2h_1^2h_2^2|}{(h_1^2 + h_2^2)^{3/2}}
L = \lim _{h \rightarrow 0} \left ||h| - \frac{2h_1^2h_2^2}{(h_1^2 + h_2^2)^{3/2}} \right |
L = 2\lim _{h \rightarrow 0} \frac{h_1^2h_2^2}{(h_1^2 + h_2^2)^{3/2}}
L = 2\lim _{h \rightarrow 0} |h| \left (\frac{h_1h_2}{h_1^2 + h_2^2} \right )^2
Now, consider the function g(z) = z + z^{-1} for positive z \in \mathbb{R}. Simple analysis shows that g reaches a minimum at 2, so:
z + \frac{1}{z} \geq 2
Now, let |h_1||h_2|^{-1} = z:
\frac{|h_1|}{|h_2|} + \frac{|h_2|}{|h_1|} \geq 2
h_1^2 + h_2^2 \geq 2|h_1||h_2|
\frac{1}{2} \geq \frac{|h_1||h_2|}{h_1^2 + h_2^2}
\frac{1}{4} \geq \left ( \frac{h_1h_2}{h_1^2 + h_2^2} \right )^2
2|h|\frac{1}{4} \geq 2|h|\left ( \frac{h_1h_2}{h_1^2 + h_2^2} \right )^2
So:
L \leq \frac{1}{2}\lim _{h \rightarrow 0} |h| = 0
Clearly, |f(h)| and |h| are non-negative, so:
0 \leq \lim _{h \rightarrow 0} \frac{|f(h)|}{|h|} = L \leq 0[/itex]<br /> <br /> This proves that f is indeed differentiable at (0, 0), and 0 (the zero transformation) is its derivative there. Q.E.D.
