Exercise 8.4 - Chapter 8 Differentials and Partial Derivatives 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $8.4$
Question 1.
Find the partial derivatives of the following functions at the indicated points.
(i) $f(x, y)=3 x^2-2 x y+y^2+5 x+2,(2,-5)$
(ii) $\mathrm{g}(\mathrm{x}, \mathrm{y})=3 \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}+2,(1,-2)$
(iii) $h(x, y, z)=x \sin (x y)+z^2 x,\left(2, \frac{\pi}{4}, 1\right)$
(iv) $G(x, y)=e^{* x+3 y} \log \left(x^2+y^2\right),(-1,1)$
Solution:
$(i)$
$
\begin{aligned}
f(x, y) & =3 x^2-2 x y+y^2+5 x+2 \\
\frac{\partial f}{\partial x} & =6 x-2 y+5, \quad \frac{\partial f}{\partial y}=-2 x+2 y \\
\text { At }(2,-5) \Rightarrow \frac{\partial f}{\partial x} & =6(2)-2(-5)+5=27 \\
\frac{\partial f}{\partial y} & =-2(2)+2(-5)=-14
\end{aligned}
$
(ii)
$
g(x, y)=3 x^2+y^2+5 x+2
$
$
\frac{\partial g}{\partial x}=6 x+5, \frac{\partial g}{\partial y}=2 y
$
$
\text { At }(1,-2) \Rightarrow \quad \begin{aligned}
\frac{\partial g}{\partial x} & =6(1)+(5)=11 \\
\frac{\partial g}{\partial y} & =2(-2)=-4
\end{aligned}
$
(iii)
$
\begin{aligned}
h(x, y, z) & =x \sin (x y)+z^2 x \\
\frac{\partial h}{\partial x} & =x[y \cos (x y)]+\sin (x y)+z^2 \\
\frac{\partial h}{\partial y} & =x^2 \cos (x y)
\end{aligned}
$

$\text { At } \begin{aligned}
\frac{\partial h}{\partial z} & =2 z x \\
\frac{\partial h}{\partial x} & \left.=2\left[\frac{\pi}{4}, 1\right) \cos \left(\frac{2 \pi}{4}\right)\right]+\sin \left(\frac{2 \pi}{4}\right)+(1)^2 \\
& =\frac{\pi}{2} \cos \left(\frac{\pi}{2}\right)+\sin \left(\frac{\pi}{2}\right)+1 \\
& =\frac{\pi}{2}(0)+1+1=2 \\
\frac{\partial h}{\partial y} & =(2)^2 \cos \left(\frac{2 \pi}{4}\right) \\
& =4 \cos \left(\frac{\pi}{2}\right) \\
& =4(0)=0 \\
\frac{\partial h}{\partial z} & =2(1)(2)=4
\end{aligned}$

(iv)
$
\begin{aligned}
\mathrm{G}(x, y) & =e^{x+3 y} \log \left(x^2+y^2\right) \\
\frac{\partial \mathrm{G}}{\partial x} & =e^{x+3 y}\left[\frac{2 x}{x^2+y^2}\right]+\log \left(x^2+y^2\right)\left[e^{x+3 y}\right] \\
& =e^{x+3 y}\left[\frac{2 x}{x^2+y^2}+\log \left(x^2+y^2\right)\right] \\
\frac{\partial \mathrm{G}}{\partial y} & =e^{x+3 y}\left[\frac{2 y}{x^2+y^2}\right]+\log \left(x^2+y^2\right)\left[3 e^{x+3 y}\right] \\
\frac{\partial \mathrm{G}}{\partial y} & =e^{x+3 y}\left[\frac{2 y}{x^2+y^2}+3 \log \left(x^2+y^2\right)\right] \\
\frac{\partial \mathrm{G}}{\partial x} & =e^{-1+3}\left[\frac{-2}{1+1}+\log (1+1)\right] \\
& =e^2[-1+\log 2]=e^2[\log 2-1] \\
\frac{\partial \mathrm{G}}{\partial y} & =e^{-1+3}\left[\frac{2}{1+1}+3 \log (1+1)\right] \\
& =e^2[1+3 \log 2]=e^2[1+\log 8]
\end{aligned}
$

Question 2.
For each of the following functions find the $f_x, f_y$ and show that $f_{x y}=f_{y x}$
(i) $f(x, y)=\frac{3 x}{y+\sin x}$
(ii) $f(x, y)=\tan ^{-1}\left(\frac{x}{y}\right)$
(iii) $f(x, y)=\cos \left(x^2-3 x y\right)$
Solution:

$\text { (i) } \begin{aligned}
f(x, y) & =\frac{3 x}{y+\sin x} \\
f_x & =\frac{(y+\sin x)[3]-3 x[0+\cos x]}{(y+\sin x)^2}=\frac{3 y+3 \sin x-3 x \cos x}{(y+\sin x)^2} \\
f_y & =\frac{(y+\sin x)[0]-3 x[1+0]}{(y+\sin x)^2}=\frac{-3 x}{(y+\sin x)^2} \\
\frac{\partial^2 f}{\partial x \partial y} & =\frac{\partial}{\partial x}\left[\frac{-3 x}{(y+\sin x)^2}\right] \\
& =\frac{(y+\sin x)^2[-3]-(-3 x) 2(y+\sin x)[0+\cos x]}{(y+\sin x)^4} \\
& =\frac{-3(y+\sin x)^2+6 x \cos x(y+\sin x)}{(y+\sin x)^4} \\
\frac{\partial^2 f}{\partial y \partial x} & =\frac{\partial}{\partial y}\left[\frac{3 y+3 \sin x-3 x \cos x]}{(y+\sin x)^2}\right] \\
& =\frac{(y+\sin x)^2[3]-(3 y+3 \sin x-3 x \cos x) 2(y+\sin x)(0+\cos x)}{(y+\sin x)^4} \\
& =\frac{-3(y+\sin x)^2+6 x \cos x(y+\sin x)}{(y+\sin x)^4} \ldots(2)
\end{aligned}$

$\text { from (1) & (2) } \Rightarrow \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial^2 f}{\partial y \partial x}$

$\begin{aligned}
& \text { (ii) } f(x, y)=\tan ^{-1}\left(\frac{x}{y}\right) \\
& \frac{\partial f}{\partial x}=\frac{1}{1+\frac{x^2}{y^2}}\left(\frac{1}{y}\right)=\frac{y}{x^2+y^2} \\
& \frac{\partial f}{\partial y}=\frac{1}{1+\frac{x^2}{y^2}} \quad\left(\frac{-x}{y^2}\right)=\frac{-x}{x^2+y^2} \\
& \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial y}\right] \\
& =\frac{\partial}{\partial x}\left[\frac{-x}{x^2+y^2}\right] \\
& =\frac{\left(x^2+y^2\right)[-1]-(-x)[2 x]}{\left(x^2+y^2\right)^2} \\
& =\frac{x^2-y^2}{\left(x^2+y^2\right)^2} \\
& \frac{\partial^2 f}{\partial y \partial x}=\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right] \\
& =\frac{\partial}{\partial y}\left[\frac{y}{x^2+y^2}\right] \\
& =\frac{\left(x^2+y^2\right)[1]-y[2 y]}{\left(x^2+y^2\right)^2} \\
& =\frac{x^2-y^2}{\left(x^2+y^2\right)^2} \\
& \text { from (1) \& (2) } \Rightarrow \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial^2 f}{\partial y \partial x} \\
&
\end{aligned}$

$\begin{aligned}
&\text { (iii) } \begin{aligned}
f(x, y) & =\cos \left(x^2-3 x y\right) \\
\frac{\partial f}{\partial x} & =-\sin \left(x^2-3 x y\right)[2 x-3 y] \\
& =(3 y-2 x) \sin \left(x^2-3 x y\right)
\end{aligned}\\
&\begin{aligned}
\frac{\partial f}{\partial y} & =-\sin \left(x^2-3 x y\right)[0-3 x] \\
& =3 x \sin \left(x^2-3 x y\right) \\
\frac{\partial^2 f}{\partial x \partial y} & =\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial y}\right] \\
& =\frac{\partial}{\partial x}\left[3 x \sin \left(x^2-3 x y\right)\right] \\
& =3 x\left[\cos \left(x^2-3 x y\right) \cdot(2 x-3 y)+\sin \left(x^2-3 x y\right)[3]\right] \\
& =3 x(2 x-3 y) \cos \left(x^2-3 x y\right)+3 \sin \left(x^2-3 x y\right) \quad \ldots .(1) \\
& =\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right] \\
& =\frac{\partial}{\partial y}\left[(3 y-2 x) \sin \left(x^2-3 x y\right)\right] \\
& =(3 y-2 x)\left[\cos \left(x^2-3 x y\right) \cdot(-3 x)\right]+\sin \left(x^2-3 x y\right)[3] \\
& =3 x(2 x-3 y) \cos \left(x^2-3 x y\right)+3 \sin \left(x^2-3 x y\right) \quad \ldots .(2) \\
\text { from (1) \& (2) } \Rightarrow \frac{\partial^2 f}{\partial x \partial y} & =\frac{\partial^2 f}{\partial y \partial x}
\end{aligned}
\end{aligned}$

Question 3.
If $\mathbf{U}(x, y, z)=\frac{x^2+y^2}{x y}+3 z^2 y$, find $\frac{\partial \mathbf{U}}{\partial x}, \frac{\partial \mathbf{U}}{\partial y}$ and $\frac{\partial \mathbf{U}}{\partial z}$

Solution:
$
\begin{aligned}
\mathrm{U}(x, y, z) & =\frac{x^2+y^2}{x y}+3 z^2 y \\
\frac{\partial \mathrm{U}}{\partial x} & =\frac{(x y)(2 x)-\left(x^2+y^2\right)(y)}{(x y)^2}+0 \\
& =\frac{2 x^2 y-x^2 y-y^3}{(x y)^2}=\frac{x^2 y-y^3}{(x y)^2} \\
& =\frac{y\left(x^2-y^2\right)}{x^2 y^2}=\frac{x^2-y^2}{x^2 y} \\
\frac{\partial \mathrm{U}}{\partial y} & =\frac{(x y)(2 y)-\left(x^2+y^2\right)(x)}{(x y)^2}+3 z^2 \\
& =\frac{2 x y^2-x^3-y^2 x}{(x y)^2}+3 z^2=\frac{x y^2-x^3}{(x y)^2}+3 z^2 \\
& =\frac{x\left(y^2-x^2\right)}{x^2 y^2}=\frac{y^2-x^2}{y^2 x}+3 z^2 \\
\frac{\partial \mathrm{U}}{\partial z} & =0+6 z y=6 z y
\end{aligned}
$
Question 4.
If $\mathrm{U}(x, y, z)=\log \left(x^2+y^3+z^3\right)$, find $\frac{\partial \mathrm{U}}{\partial x}+\frac{\partial \mathrm{U}}{\partial y}$ and $\frac{\partial \mathrm{U}}{\partial z}$.

Solution:
$
\begin{aligned}
\mathrm{U}(x, y, z) & =\log \left(x^3+y^3+z^3\right) \\
\frac{\partial U}{\partial x} & =\frac{3 x^2}{x^3+y^3+z^3} \\
\frac{\partial U}{\partial y} & =\frac{3 y^2}{x^3+y^3+z^3} \\
\frac{\partial U}{\partial z} & =\frac{3 z^2}{x^3+y^3+z^3} \\
\frac{\partial U}{\partial x}+\frac{\partial U}{\partial y}+\frac{\partial U}{\partial z} & =\frac{3\left(x^2+y^2+z^2\right)}{\left(x^3+y^3+z^3\right)}
\end{aligned}
$
Question 5.
For each of the following functions find the $g_{x y}, g_{x x}, g_{y y}$ and $g_{y x}$.
(i) $g(x, y)=x_y+3 x_2 y$
(ii) $g(x, y)=\log (5 x+3 y)$
(iii) $g(x, y)=x_2+3 x y-7 y+\cos (5 x)$
Solution:

(i)
$
\begin{aligned}
\frac{\partial g}{\partial x} & =g_x=e^y+6 x y \\
\frac{\partial g}{\partial y}= & g_y=x e^y+3 x^2 \\
g_{x x}=\frac{\partial^2 g}{\partial x^2} & =\frac{\partial}{\partial x}\left[\frac{\partial g}{\partial x}\right]=\frac{\partial}{\partial x}\left[e^y+6 x y\right] \\
& =0+6 y=6 y \\
g_{y y}=\frac{\partial^2 g}{\partial y^2} & =\frac{\partial}{\partial y}\left[\frac{\partial g}{\partial y}\right]=\frac{\partial}{\partial y}\left[x e^y+3 x^2\right] \\
& =x e^y \\
g_{x y}=\frac{\partial^2 g}{\partial x \partial y} & =\frac{\partial}{\partial x}\left[\frac{\partial g}{\partial y}\right]=\frac{\partial}{\partial x}\left[x e^y+3 x^2\right] \\
& =e^y+6 x \\
g_{y x}=\frac{\partial^2 g}{\partial y \partial x} & =\frac{\partial}{\partial y}\left[\frac{\partial g}{\partial x}\right]=\frac{\partial}{\partial y}\left[e^y+6 x y\right] \\
& =e^y+6 x
\end{aligned}
$

(ii)
$
\begin{aligned}
& g_x=\frac{\partial g}{\partial x}=\frac{1}{5 x+3 y}(5)=\frac{5}{5 x+3 y} \\
& g_y=\frac{\partial g}{\partial y}=\frac{1}{5 x+3 y}(3)=\frac{3}{5 x+3 y}
\end{aligned}
$

$\begin{aligned}
g_{x x}=\frac{\partial^2 g}{\partial x^2} & =\frac{\partial}{\partial x}\left[\frac{\partial g}{\partial x}\right]=\frac{\partial}{\partial x}\left[\frac{5}{5 x+3 y}\right] \\
& =\frac{(5 x+3 y)(0)-5(5)}{(5 x+3 y)^2}=\frac{-25}{(5 x+3 y)^2} \\
g_{y y}=\frac{\partial^2 g}{\partial y^2} & =\frac{\partial}{\partial y}\left[\frac{\partial g}{\partial y}\right]=\frac{\partial}{\partial y}\left[\frac{3}{5 x+3 y}\right] \\
& =\frac{(5 x+3 y)(0)-3(3)}{(5 x+3 y)^2} \\
& =\frac{-9}{(5 x+3 y)^2} \\
g_{x y}=\frac{\partial^2 g}{\partial x \partial y} & =\frac{\partial}{\partial x}\left[\frac{\partial g}{\partial y}\right]=\frac{\partial}{\partial x}\left[\frac{3}{5 x+3 y}\right] \\
& =\frac{-3}{(5 x+3 y)^2}(5) \\
& =\frac{-15}{(5 x+3 y)^2}
\end{aligned}$

(iii)
$
\begin{aligned}
g(x, y) & =x^2+3 x y-7 y+\cos (5 x) \\
g_x & =\frac{\partial g}{\partial x}=2 x+3 y-5 \sin (5 x) \\
g_y & =\frac{\partial g}{\partial y}=3 x-7 \\
g_{x x} & =2-25 \cos (5 x) \\
g_{y y} & =0 \\
g_{x y} & =\frac{\partial}{\partial x}\left(\frac{\partial g}{\partial y}\right)=\frac{\partial}{\partial x}(3 x-7)=3 \\
g_{y x} & =\frac{\partial}{\partial y}\left(\frac{\partial g}{\partial x}\right)=\frac{\partial}{\partial y}(2 x+3 y-5 \sin (5 x))=3
\end{aligned}
$

Question 6.
Let $w(x, y, z)=\frac{1}{\sqrt{x^2+y^2+z^2}}(x, y, z) \neq(0,0,0)$. Show that $\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}+\frac{\partial^2 w}{\partial z^2}=0$
Solution:

$\begin{aligned}
& w(x, y, z)=\frac{1}{\sqrt{x^2+y^2+z^2}} \\
& \frac{\partial w}{\partial x}=\frac{-\frac{1}{2}(2 x)}{\left(x^2+y^2+z^2\right)^{3 / 2}}=\frac{-x}{\left(x^2+y^2+z^2\right)^{3 / 2}} \\
& \frac{\partial w}{\partial y}=\frac{-y}{\left(x^2+y^2+z^2\right)^{3 / 2}} \\
& \frac{\partial^2 w}{\partial x^2}=\frac{\left(x^2+y^2+z^2\right)^{3 / 2}(-1)+x \frac{3}{2}\left(x^2+y^2+z^2\right)^{1 / 2}(2 x)}{\left[\left(x^2+y^2+z^2\right)^{3 / 2}\right]^2} \\
& =\frac{\left(x^2+y^2+z^2\right)^{1 / 2}\left[-x^2-y^2-z^2+3 x^2\right]}{\left(x^2+y^2+z^2\right)^3} \\
& =\frac{2 x^2-y^2-z^2}{\left(x^2+y^2+z^2\right)^{\frac{5}{2}}} \\
& \frac{\partial^2 w}{\partial y^2}=\frac{-x^2+2 y^2-z^2}{\left(x^2+y^2+z^2\right)^{\frac{5}{2}}} \\
& \frac{\partial^2 w}{\partial z^2}=\frac{-x^2-y^2+2 z^2}{\left(x^2+y^2+z^2\right)^{\frac{5}{2}}} \\
&
\end{aligned}$

$\begin{aligned}
\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}+\frac{\partial^2 w}{\partial z^2} & =\frac{0}{\left(x^2+y^2+z^2\right)^{\frac{5}{2}}} \\
\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}+\frac{\partial^2 w}{\partial z^2} & =0
\end{aligned}$

Question 7.
If $\mathrm{V}(x, y)=e^x(x \cos y-y \sin y)$, then prove that $\frac{\partial^2 \mathrm{~V}}{\partial x^2}+\frac{\partial^2 \mathrm{~V}}{\partial y^2}=0$
Solution:
$
\begin{aligned}
\mathrm{V}(x, y) & =e^x[x \cos y-y \sin y] \\
\mathrm{V} & =x e^x \cos y-e^x y \sin y \\
\frac{\partial \mathrm{V}}{\partial x} & =\cos y\left[e^x+x e^x\right]-e^x y \sin y \\
\frac{\partial^2 \mathrm{~V}}{\partial x^2} & =\cos y\left[e^x+e^x+x e^x\right]-e^x y \sin y \\
\frac{\partial^2 \mathrm{~V}}{\partial x^2} & =e^x \cos y+e^x \cos y+x e^x \cos y-e^x y \sin y \\
\frac{\partial \mathrm{V}}{\partial y} & =x e^x[-\sin y]-e^x[y \cos y+\sin y] \\
\frac{\partial^2 \mathrm{~V}}{\partial y^2} & =-x e^x \cos y-e^x[-y \sin y+\cos y+\cos y] \\
\frac{\partial^2 \mathrm{~V}}{\partial y^2} & =-x e^x \cos y+e^x y \sin y-e^x \cos y-e^x \cos y \\
(1)+(2) & \Rightarrow \frac{\partial^2 \mathrm{~V}}{\partial x^2}+\frac{\partial^2 \mathrm{~V}}{\partial y^2}=0
\end{aligned}
$

Question 8.
If $w(x, y)=x y+\sin (x y)$, then prove that $\frac{\partial^2 w}{\partial y \partial x}=\frac{\partial^2 w}{\partial x \partial y}$
Solution:
$
\begin{aligned}
\mathrm{W}(x, y) & =x y+\sin (x y) \\
\text { LHS } & =\frac{\partial^2 w}{\partial y \partial x} \\
& =\frac{\partial}{\partial y}\left[\frac{\partial w}{\partial x}\right] \\
& =\frac{\partial}{\partial y}[y+y \cos (x y)] \\
& =1+y[-x \sin (x y)] \\
& =1-x y \sin (x y) \\
\text { RHS } & =\frac{\partial^2 w}{\partial x \partial y} \\
& =\frac{\partial}{\partial x}\left[\frac{\partial w}{\partial y}\right] \\
\text { from (1) & (2) } & \Rightarrow \frac{\partial^2 w}{\partial y \partial x}=\frac{\partial^2 w}{\partial x \partial y}
\end{aligned}
$

Question 9.
If $v(x, y, z)=x^3+y^3+z^3+3 x y z$, show that $\frac{\partial^2 v}{\partial y \partial z}=\frac{\partial^2 v}{\partial z \partial y}$

Solution:
$
\begin{aligned}
v(x, y, z)=x^3+y^3 & +z^3+3 x y z \\
\frac{\partial^2 v}{\partial y \partial z} & =\frac{\partial}{\partial y}\left[\frac{\partial v}{\partial z}\right] \\
& =\frac{\partial}{\partial y}\left[3 z^2+3 x y\right] \\
& =3 x \\
\frac{\partial^2 v}{\partial z \partial y} & =\frac{\partial}{\partial z}\left[\frac{\partial v}{\partial y}\right] \\
& =\frac{\partial}{\partial z}\left[3 y^2+3 x z\right] \\
& =3 x
\end{aligned}
$
from (1) \& (2) $\Rightarrow \frac{\partial^2 v}{\partial y \partial z}=\frac{\partial^2 v}{\partial z \partial y}$

Question 10.
A firm produces two types of calculators each week, $x$ number of type $A$ and $j$ number of type $B$. The weekly revenue and cost functions (in rupees) are $R(x, y)=80 x+90 y+0.04 x y-0.05 \cdot x^2-0.05 y^2$ and $C(x, y)=8 x+6 y+2000$ respectively.
(i) Find the profit function $\mathrm{P}(\mathrm{x}, \mathrm{y})$,
(ii) Find $\frac{\partial P}{\partial x}(1200,1800)$ and $\frac{\partial P}{\partial y}(1200,1800)$ and interpret these results.
Solution:
$
\begin{aligned}
& \text { (i) Profit = Revenue }-\text { Cost } \\
& =\left(80 x+90 y+0.04 x y-0.05 x^2-0.05 y^2\right)-(8 x+6 y+2000) \\
& =80 x+90 y+0.04 x y-0.05 x^2-0.05 y^2-8 x-6 y-2000 \\
& P(x, y)=72 x+84 y+0.04 x y-0.05 x^2-0.05 y^2-2000
\end{aligned}
$

(ii) $\begin{aligned}
\frac{\partial \mathrm{P}}{\partial x} & =72+0.04 y-0.1 x \\
\frac{\partial \mathrm{P}}{\partial y} & =84+0.04 x-0.1 y \\
\text { At }(1200,1800) \frac{\partial \mathrm{P}}{\partial x} & =72+0.04(1800)-0.1(1200) \\
\text { At }(1200,1800) & =72+72-120=24 \\
\frac{\partial \mathrm{P}}{\partial x} & =72+0.04(1800)-0.1(1200) \\
\frac{\partial \mathrm{P}}{\partial y} & =72+72-120=24 \\
& =84+0.04(1200)-0.1(1800) \\
& 84+48-180=-48
\end{aligned}$