Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.58 of these notes.

# Question 1

## (i)

$\frac{\partial f}{\partial x}=3x^2-4y^2(1)+0$.

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$\frac{\partial f}{\partial y}=-4x(2y)+4y^3=4y^4-8xy$.

## (ii)

$\frac{\partial f}{\partial x}=e^y(2x)+0=2xe^y$.

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$\frac{\partial f}{\partial y}=x^2(e^y)-4=x^2e^y-4$.

## (iii)

This one needs a product and a chain rule for $f_x$ and a chain rule for $f_y$.

$\frac{\partial f}{\partial x}=x^2\times\frac{\partial \sin xy}{\partial x}+\sin xy\times\frac{\partial x^2}{x}+0$

$=x^2\times \cos xy\times \frac{\partial xy}{\partial x}+\sin xy\times 2x$

$=x^2\cos xy \times y+2x\sin xy=x^2y\cos xy+2x \sin xy$.

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$\frac{\partial f}{\partial y}=x^2\times \cos xy\times\frac{\partial xy}{\partial y}-6y$

$=x^2\times\cos xy\times x-6y=x^3\cos xy-6y$.

## (iv)

$\frac{\partial f}{\partial x}=\sin y(3)+4y^2z(3x^2)=3(\sin y+4x^2y^2z)$.

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$\frac{\partial f}{\partial y}=3x(\cos y)+4x^3z(2y)=3x\cos y+8x^2yz$.

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$\frac{\partial f}{\partial z}=0+4x^3y^2(1)=4x^3y^2$.

# Question 2

## (i)

$f_x=3x^2-4y^2(1)+0=3x^2-4y^2$.

$f_{xx}=6x+0=6x$.

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$f_{y}=0-4x(2y)+3=-8xy+3$.

$f_{yy}=-8x(1)+0=-8x$.

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$f_{xy}=(f_{x})_y=0-8y=-9y$.

## (ii)

$f_x=4x^3-3y^3(2x)+0=4x^3-6xy^3=$.

$f_{xx}=12x^2-6y^3(1)=6(2x^2-y^3)$.

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$f_{xy}=(f_{x})_y=0-6x(3y^2)=-18xy^2$.

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$f_{xyy}=(f_{xy})_y=-18x(2y)=-36y$.

## (iii)

This example is going to require the chain rule when we differentiate $e^{2xy}$. Also to ease differentiation, we use the fact that $1/a^n=a^{-n}$ to write:

$f(x,y,z)=e^{2xy}-z^2y^{-1}+xz\sin y$.

$f_x=e^{2xy}\times\frac{\partial 2xy}{\partial x}+0+z\sin y =e^{2xy}\times 2y+z\sin y=2ye^{2xy}+z\sin y$.

$f_{xx}=2ye^{2xy}\times\frac{\partial 2xy}{\partial y}+0=2ye^{2xy}\times 2y=4y^2e^{2xy}$.

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$f_y=e^{2xy}\times\frac{\partial 2xy}{\partial y}-z^2(-1y^{-2})+xz(\cos y)=e^{2xy}\times 2x+z^2y^{-2}+xz\cos y$,

$=2xe^{2xy}+z^2y^{-2}+xz\cos y$.

$f_{yy}=2xe^{2xy}\frac{\partial 2xy}{\partial y}+z^2((-2)y^{-3})+xz(-\sin y)$,

$2xe^{2xy}\times 2x-2\frac{z^2}{y^3}-xz\sin y=4x^2e^{2xy}-2\frac{z^2}{y^3}-xz\sin y$.

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Well $f_{yyzz}=(f_{yy})_{zz}$ so first we evaluate:

$(f_{yy})_z=0-\frac{4z}{y^3}(1)-x\sin y(1)=\frac{4z}{y^3}-x\sin y$,

$(f_{yy})_{zz}=\frac{4}{y^3}$.