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Andresfon12

  • 2 years ago

find the partial derivatives

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  1. Andresfon12
    • 2 years ago
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    \[a) \frac{ z }{ y }| (1, 1/2) if z= e^{x+2y} \sin y\] b) fx (pi/3, 1) if f(x, y) = x ln (y cos x)

  2. ZeHanz
    • 2 years ago
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    a) I don't understand, maybe something went wrong with typing in the formula? b) You need fx, so consider y as a constant. e'll need the Product Rule here, and also the Chain Rule...

  3. Andresfon12
    • 2 years ago
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    |dw:1361122439795:dw|

  4. Andresfon12
    • 2 years ago
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    fx = 1/ (cos x)* -sin (x)

  5. muhammad9t5
    • 2 years ago
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    take ln on both sides.

  6. Andresfon12
    • 2 years ago
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    @muhammad9t ibyt i think i already did that for part b

  7. ZeHanz
    • 2 years ago
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    I understand what you wrote in a) now ;) \[\frac{ \delta z }{ \delta y }=2e^{x+2y}\sin y+e^{x+2y} \cos y=e^{x+2y}(2\sin y + \cos y)\]Now set x = 1 and y = ½: \[\frac{ \delta z }{ \delta y }(1,\frac{ 1 }{ 2 })=e^{2}(2\sin 1+\cos \frac{ 1 }{ 2 })\]

  8. ZeHanz
    • 2 years ago
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    Second one: f(x, y) = x ln(ycos x). I need fx, so y is constant:\[\frac{ \delta f }{ \delta x }=1 \cdot \ln(y \cos x)+x \cdot \frac{ 1 }{ y \cos x }\cdot -y \sin x=\ln(y \cos x)- x \tan x\]Now set x = pi/3, y = 1:\[\ln(\cos \frac{ \pi }{ 3 })-\frac{ \pi }{ 3 }\tan \frac{ \pi }{ 3 }=\ln \frac{ 1 }{ 2 }-\frac{ \pi }{ 3 }\sqrt{3}=-(\ln2+\frac{ \pi }{ 3 }\sqrt{3})\]

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