z is ahomogeneous function of degree n
\( \therefore x \frac{\partial z}{\partial x}+\mathrm{y} \frac{\partial z}{\partial y}=\mathrm{nz} \) ......(1)
\( \mathrm{V}=\mathrm{z} \mathrm{e}^{\mathrm{ax}+\mathrm{by}} \)
\( \frac{\partial V}{\partial x}=\mathrm{z} \mathrm{e}^{\mathrm{ax}+b \mathrm{~b}} \mathrm{a}+\mathrm{e}^{\mathrm{ax}+\mathrm{by} \frac{\partial z}{\partial y}} \)
\( \mathrm{x} \frac{\partial V}{\partial x}=\mathrm{e}^{a \mathrm{x}+\mathrm{by}}\left(a x z+x \frac{\partial z}{\partial x}\right) \)
\( \frac{\partial V}{\partial y}=\mathrm{ze}^{\mathrm{ax}+b \mathrm{~b}} \mathrm{~b}+\mathrm{e}^{\mathrm{ax}+\mathrm{by} \frac{\partial z}{\partial y}} \)
\( \mathrm{y} \frac{\partial V}{\partial y}=\mathrm{e}^{\mathrm{ax}+\mathrm{by}}\left(b y z+y \frac{\partial z}{\partial y}\right) \)
\( \mathrm{x} \frac{\partial V}{\partial x}+\mathrm{y} \frac{\partial V}{\partial y} =\mathrm{e}^{\mathrm{ax}+\mathrm{by}}\left(a x z+b y z+x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}\right) \\ \)
\( =\mathrm{e}^{\mathrm{ax}+\mathrm{by}}(\mathrm{axz}+\mathrm{byz}+\mathrm{nz}) \)
\( =\mathrm{z} \mathrm{e}^{\mathrm{ax}+\mathrm{by}}(\mathrm{ax}+\mathrm{by}+\mathrm{n}) \)
\( =\mathrm{V}(\mathrm{ax}+\mathrm{by}+\mathrm{n}) \)
\( \therefore \mathrm{x} \frac{\partial V}{\partial x}+\mathrm{y} \frac{\partial V}{\partial y}=\mathrm{V}(\mathrm{ax}+\mathrm{by}+\mathrm{n})\)

We know
\( \frac{d w}{d t}=\frac{\partial w}{\partial x} \frac{d x}{d t}+\frac{\partial w}{\partial y} \frac{d y}{d t}+\frac{\partial w}{\partial z} \frac{d z}{d t} \)
\(\frac{\partial w}{\partial x}=1 \quad \frac{d x}{d t}=-\sin t \)
\(\frac{\partial w}{\partial y}=2 \quad \frac{d y}{d t}=\cos t \)
\(\frac{\partial w}{\partial z}=2 \mathrm{z} \quad \frac{d z}{d y}=1 \)
\(\frac{d w}{d t}=1(-\sin t)+2 \cos t+2 \mathrm{z} \)
\(=-\sin \mathrm{t}+2 \cos \mathrm{t}+2 \mathrm{t}\)

We know
\( \frac{\partial w}{\partial x}=\frac{\partial w}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial w}{\partial v} \frac{\partial v}{\partial x} \text { and } \)
\(\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial w}{\partial v} \frac{\partial v}{\partial y} \)
\(\mathrm{w}=\mathrm{u}^{2} \mathrm{e}^{\mathrm{v}}\)
\(\frac{\partial w}{\partial u}=2 \mathrm{u} \mathrm{e}^{v} \quad \frac{\partial w}{\partial v}=u^{2} \mathrm{e}^{v} \)
\(\frac{\partial u}{\partial x}=\frac{1}{y} \quad \frac{\partial u}{\partial y}=-\frac{x}{y^{2}}\)
\(\frac{\partial v}{\partial x}=\frac{y}{x} \ \frac{\partial u}{\partial y}=-\frac{x}{y^{2}}  \)
\(\frac{\partial w}{\partial x}=\frac{2 u e^{v}}{y}+u^{2} e^{v} \frac{y}{x}\)
\(\frac{\partial w}{\partial x}=\frac{2 u e^{v}}{y}+u^{2} e^{v} \frac{y}{x}\)
\( =u e^{y}\left(\frac{2}{y}+u \frac{y}{x}\right) \)
\( =\frac{x}{y} e^{y \log x}\left(\frac{2}{y}+\frac{\not x}{\not p} \cdot \frac{\not p}{\not x}\right) \)
\(=\frac{x}{y} x^{y}\left(\frac{2}{y}+1\right) \)
\( =x^{y} \frac{x}{y^{2}}(2+y) \)
\( \frac{\partial w}{\partial y} =2 \mathrm{u} \mathrm{}^{v}-\frac{x}{y^{2}}+\mathrm{u}^{2} \mathrm{e}^{\mathrm{v}} \log \mathrm{x} \)
\( =u e^{v}\left(\frac{-2 x}{y^{2}}+u \log x\right) \)
\( =\frac{x}{y} e^{y \log x}\left(\frac{-2 x}{y^{2}}+\frac{x}{y} \log x\right) \)
\(\text {Since } \mathrm{u}=\frac{x}{y}, \mathrm{v}=\mathrm{y} \log \mathrm{x} \)
\( =\frac{x}{y} x^{y}\left(\frac{-2 x+x y}{y^{2}} \log x\right) \)
\(=\frac{x^{2}}{y^{3}} x^{y}[y \log x-2] \)

\(\mathrm{Z}=y e^{x^{2}}\) where x = 2t and y = 1-t
\( \frac{\partial Z}{\partial x}=y e^{x^{2}} 2 \mathrm{x} \)
\(\frac{\partial Z}{\partial y}=e^{x^{2}} \)
\(\frac{\partial x}{\partial t}=2 \)
\(\frac{\partial y}{\partial t}=-1 \)
\(\frac{d Z}{d t}=\frac{\partial Z}{\partial x} \frac{\partial x}{\partial t}+\frac{\partial Z}{\partial y} \frac{\partial y}{\partial t} \)
\( =y e^{x^{2}} 2 x(2)+e^{x^{2}}(-1) \)
\(=4 x y e^{x^{2}}-e^{x^{2}} \)
\(=e^{4 r^{2}}(8 t(1-t)-1) \)
Since x = 2t, y = 1-t
\(=e^{4 t^{2}}\left(8 t-8 t^{2}-2\right)\)

\( f(x, y) =\frac{1}{\sqrt{x^{2}+y^{2}}} \)
\(f(\lambda x, \lambda y) =\frac{1}{\sqrt{\lambda^{2} x^{2}+\lambda^{2} y^{2}}}=\frac{1}{\lambda \sqrt{x^{2}+y^{2}}} \)
f is a homogeneous function of degree -1 
x- y By Eulers Theorem
\(x \frac{\partial f}{\partial x}+y \frac{\partial u}{\partial y}=-\mathrm{f}\)
Verification:
\( \mathbf{f} =\left(x^{2}+y^{2}\right)^{-1 / 2} \)
\(\frac{\partial f}{\partial x} =-\frac{1}{2}\left(x^{2}+y^{2}\right)^{-1 / 2} \times 2 x\)
\( =-\frac{x}{\left(x^{2}+y^{2}\right)^{3 / 2}} \)
\(\mathrm{x} \frac{\partial f}{\partial x} =-\frac{x^{2}}{\left(x^{2}+y^{2}\right)^{1 / 2}} \)
Similarly,
\( \mathrm{y} \frac{\partial f}{\partial y} =-\frac{y^{2}}{\left(x^{2}+y^{2}\right)^{3 / 2}} \)
\(\mathrm{x} \frac{\partial f}{\partial x}+\mathrm{y} \frac{\partial u}{\partial y} =-\frac{\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3 / 2}} \)
\( =-\frac{1}{\sqrt{x^{2}+y^{2}}}=-\mathrm{f}\)

\( \mathrm{u} =\sin ^{-1}\left(\frac{x-y}{\sqrt{x}+\sqrt{y}}\right) \)
\(\sin \mathrm{u} =\left(\frac{x-y}{\sqrt{x}+\sqrt{y}}\right)=\mathrm{f}(\mathrm{x}, \mathrm{y}) \)
\( f(\lambda x, \lambda y) =\frac{\lambda x-\lambda y}{\sqrt{\lambda x}+\sqrt{\lambda y}} \)
\( =\frac{\lambda(x-y)}{\sqrt{\lambda}(\sqrt{x}+\sqrt{y})} \)
\( =\lambda^{1 / 2} \frac{(x-y)}{(\sqrt{x}+\sqrt{y})} \)
f is a homogeneous function of degree \(\frac{1}{2}\) By Eulers Theorem
\( \mathrm{x} \frac{\partial f}{\partial x}+\mathrm{y} \frac{\partial u}{\partial y} =\frac{1}{2} \mathrm{f} \)
\(\mathrm{x} \frac{\partial}{\partial x} \sin u+\mathrm{y} \frac{\partial}{\partial y} \sin \mathrm{u} =\frac{1}{2} \sin u \)
\(\mathrm{x} \cos \mathrm{u} \frac{\partial u}{\partial x}+\mathrm{y} \cos \mathrm{u} \frac{\partial u}{\partial y} =\frac{1}{2} \sin \mathrm{u} \)
\(\mathrm{x} \frac{\partial u}{\partial x}+\mathrm{y} \frac{\partial u}{\partial y} =\frac{1}{2} \frac{\sin u}{\cos u} \)
\(\mathrm{x} \frac{\partial u}{\partial x}+\mathrm{y} \frac{\partial u}{\partial y} =\frac{1}{2} \tan \mathrm{u}\)

\( f(x)=(27)^{1 / 3}, \Delta x=-2 \)
\( f(x+\Delta x)=f(x)+f^{\prime}(x) \Delta x \)
\( =3+\frac{1}{3 x^{2 / 3}} x-2 \)
\( =3-\frac{2}{27}\)
= 3 - 0.074
\((25)^{1 / 3}=2.926\)
Which is the required approximate value

\( f(x)=\log _{10} x, \text { then } f^{\prime}(x)=\frac{1}{x} \log _{10} e \)
\( f(x+\Delta x)=f(x)+f^{\prime}(x) \Delta x \)
\( \log 10.1=\log _{10} 10+\frac{0.4343}{10} \times 0.1\)
log10.1 = 1+0.004343 = 1.004343
= 1.004343
Hence, the approximate value of
log₁₀10.1 = 1.004343

2.0116 approx.

\( \frac { \partial w }{ \partial u } =\frac { 2u }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } ;\frac { \partial w }{ \partial v } =\frac { -2v }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } \)

Given z = f(x - cy) + F (x + cy)
\(\frac { \partial z }{ \partial x } \) = f'(x - cy) (1) + F'(x + cy) (1)
= f'(x - cy) + F'(x + cy)
\(\frac { { \partial }^{ 2 }z }{ \partial { x }^{ 2 } } =\frac { \partial }{ \partial x } \left( \frac { \partial z }{ \partial x } \right) \)
= f"(x - cy) + F"(x + cy) ... (1)
\(\frac { \partial z }{ \partial y } \) = f'(x - cy)(-c) + F'(x + cy)(c)
\(\frac { { \partial }^{ 2 }z }{ \partial { y }^{ 2 } } \) = f"(x - cy).(c²) + F"(x + cy)(c²)
= c² [f"(x - cy) + F"(x + cy)] ... (2)
\(\frac { { \partial }^{ 2 }z }{ \partial { y }^{ 2 } } ={ c }^{ 2 }\frac { { \partial }^{ 2 }z }{ \partial { x }^{ 2 } } \)   [using (1)]

Given r² = x² +y² + z²
log r² = log (x² + y² + z²)
⇒ 2 log r = log (x² + y² + z²)
∴ 2V = log (x² + y² + z²) [∵ V = log r ]
⇒ V = \(\frac12\) log (x² + y² + z²)
\(\frac { \partial V }{ \partial x } =\frac { 1 }{ 2 } \frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } =\frac { x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } =\frac { ({ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 })(1)-x(2x) }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-{ x }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
III^ty \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ y }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
∴ \(\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(\therefore \frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { y }^{ 2 }+{ z }^{ 2 }-{ x }^{ 2 }+{ z }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } \)
= \(\frac { { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } } \)
= \(\frac { 1 }{ { r }^{ 2 } } \)
Hence proved.

Let f(x) = cos x, x = 60° dx = 1°
f(x_o) = cos 60° = \(\frac12\) = 0.5
f'(x) = - sinx dx
f'(x_o) = - sin x_o dx
f'(60) = - sin 60° (1°)
= - (0.86603) (0.01745)
= - 0.0154
∴ f(x) = f(x_o) +f(x_o) dx 
f(61) = 0.5 - 0.0154
∴ tan 46° = f(x_o) +f(x_o) dx 
cos 61° = 0.4849

Given f(x, y) = 2x² + 3y² - 2xy
\(\frac { \partial f }{ \partial x } \) = 4x - 8y
\({ \left( \frac { \partial f }{ \partial x } \right) }_{ (2,3) }\) = 4(2) - 8(3)
= 8 - 24 = -16
\(\frac { \partial f }{ \partial y } \) = 6y-8x
\({ \left( \frac { \partial f }{ \partial y } \right) }_{ (2,3) }\) = 6(3)- 8(2)
= 18-16 = 2
\(\frac { { \partial }^{ 2 }f }{ { \partial x }^{ 2 } } =\frac { \partial }{ \partial x } { \left( \frac { \partial f }{ \partial x } \right) }=4\)
\(\frac { { \partial }^{ 2 }f }{ { \partial y }^{ 2 } } =\frac { \partial }{ \partial y } { \left( \frac { \partial f }{ \partial y } \right) }=6\)

Given u = tan^-1 \(\left( \frac { { x }^{ 3 }+{ y }^{ 3 } }{ x-y } \right) \)
⇒ tan u \(\frac { { x }^{ 3 }+{ y }^{ 3 } }{ x-y } \) and
let f = tan u
∴ f(x,y) = \(\frac { { x }^{ 3 }+{ y }^{ 3 } }{ x-y } \)
\(f(tx,ty)=\frac { { t }^{ 3 }{ x }^{ 3 }+{ t }^{ 2 }{ y }^{ 3 } }{ tx-ty } =\frac { { t }^{ 3 }({ x }^{ 3 }+{ y }^{ 3 }) }{ t(x-y) } \)
= t²f(x, y)
∴ f(x, y) is a homogeneous function of degree 2.
∴ By Euler's theorem,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 2f
⇒ \(x.\frac { \partial }{ \partial x } \) (tan u) + y.\(\frac { \partial }{ \partial y } \) (tan u)
= 2 tan u [∵ f = tan u]
⇒ x sec² u\(\frac { \partial u }{ \partial x } \) + y sec² u \(\frac { \partial u }{ \partial y } \) = 2 tan u
Dividing by sec² u we get,
\(x\frac { \partial u }{ \partial y } +y\frac { \partial u }{ \partial y } =\frac { 2sinu }{ { sec }^{ 2 }u } =\frac { 2sinu }{ \frac { 1 }{ { cos }^{ 2 }u } } \)
= 2 sin u cos u = sin 2 u
Hence proved.