Partial Derivatives MCQ Quiz in मल्याळम - Objective Question with Answer for Partial Derivatives - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

A function f(x, y) is said to be homogeneous of degree n in x and y, it can be written in the form

f(λx, λy) = λn f(x, y)

Euler’s theorem:

If f(x, y) is a homogeneous function of degree n in x and y and has continuous first and second-order partial derivatives, then

$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$

$x^2\frac{{\partial }^{2}f}{\partial x^2}+2xy\frac{{\partial }^{2}f}{\partial x\partial y}+y^2\frac{\partial f}{\partial y}=n\left(n-1\right)f$

If z is homogeneous function of x & y of degree n and z = f(u), then

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=n\frac{f\left(u\right)}{f^{\prime }\left(u\right)}$

Calculation:

Given, $u={\log }_e\left(\frac{x^4+y^4}{x+y}\right)$

$z=\frac{x^4+y^4}{x+y}$

z is a homogenous function of x & y with a degree 3.

Now, z = eu

Thus, by Euler’s theorem:

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=3\frac{e^u}{e^u}$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=3$

f(x, y, z) = (4x³y – 3y²x² + 4z) lny

$fy=\frac{\partial f}{\partial y}=\left(4x^3-6y{x}^2\right)\ln y+\left(4x^{3}y-3y^2{x}^2+4z\right)\frac{1}{y}$

At point x = 2, y = 1, z = 2

f_y = 0 + (4(8)(1) - 3(1)(4) + 4(2))

= 28

$T=2\pi \sqrt{\frac{l}{g}}$

$\log T=\log 2\pi +\frac{1}{2}\log l-\frac{1}{2}\log g$

$\frac{1}{T}\delta T=0+\frac{1}{2}\frac{\delta l}{l}-\frac{1}{2}\frac{\delta g}{g}$

$\frac{\delta T}{T}100=\frac{1}{2}\left(\frac{\delta l}{l}100-\frac{\delta g}{g}100\right)$

$=\frac{1}{2}\left(1\pm 2.5\right)$

= 1.75% or -0.75%

Target is having inclination of 45° with x-axis

$\begin{array}{l}\Rightarrow \frac{dy}{dx}=\tan {45}^{\circ }\\ \Rightarrow \frac{d\left(xlnx\right)}{dx}=1\\ \Rightarrow \ln x+\frac{x}{x}=1\end{array}$

$\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=\mathrm{x}\mathrm{y}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}$

$\frac{\partial }{\partial \mathrm{y}}\left(\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right)=\frac{\partial }{\partial \mathrm{y}}(\mathrm{x}\mathrm{y}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}})$

$\frac{{\partial }^2\mathrm{u}}{\partial \mathrm{y}\partial \mathrm{z}}=\mathrm{x}\mathrm{y}\frac{\partial }{\partial \mathrm{y}}({\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}})+{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}\frac{\partial }{\partial \mathrm{y}}(\mathrm{x}\mathrm{y})$

= xy(xz)e^xyz + xe^xyz

$\frac{\partial }{\partial \mathrm{x}}\left(\frac{{\partial }^2\mathrm{u}}{\partial \mathrm{y}\partial \mathrm{z}}\right)=\frac{\partial }{\partial \mathrm{x}}({\mathrm{x}}^2\mathrm{y}\mathrm{z}+\mathrm{x}){\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}$

$\frac{{\partial }^3\mathrm{u}}{\partial \mathrm{x}\partial \mathrm{y}\partial \mathrm{z}}=({\mathrm{x}}^2\mathrm{y}\mathrm{z}+\mathrm{x})\mathrm{y}\mathrm{z}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}+{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}(2\mathrm{x}\mathrm{y}\mathrm{z}+1)$

$={\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}({\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2+\mathrm{x}\mathrm{y}\mathrm{z}+2\mathrm{x}\mathrm{y}\mathrm{z}+1)$

= (1 + 3xyz + x²y²z²) e^xyz

putting x, y, z = 1, 1, 1, we get

$\frac{{\partial }^3\mathrm{u}}{\partial \mathrm{x}\partial \mathrm{y}\partial \mathrm{z}}=(1+3+1)\mathrm{e}$

= 5e

Concept:

$\frac{\partial H}{\partial v}=\frac{\partial H}{\partial t}\times \frac{\partial t}{\partial v}$

Calculation:

Given: 

$\begin{array}{l}H={\tan }^{-1}\frac{x}{y}\end{array}$

x = u + v

y = u - v

substituting values of x and y in H, we get
$H={\tan }^{-1}\left[\frac{u+v}{u-v}\right]$

let $t=\frac{u+v}{u-v}$------------(1)

$H={\tan }^{-1}t$

differentiating w.r.t 't'

$\begin{array}{l}\frac{\partial H}{\partial t}=\frac{1}{1+t^2}\end{array}$------------(2)

In equation (1), differentiate 't' w.r.t v

$\frac{\partial t}{\partial v}=\frac{(u-v)(1)-(u+v)(-1)}{{\left(u-v\right)}^2}=\frac{2u}{{\left(u-v\right)}^2}$---------(3)

Multiplying (2) and (3), we get

$\frac{\partial H}{\partial v}=\frac{1}{1+t^2}\left[\frac{2u}{{\left(u-v\right)}^2}\right]$

By substituting '$t=\frac{u+v}{u-v}$' we get

$\frac{\partial H}{\partial v}=\frac{u}{u^2+v^2}$

Taking the partial derivative on both the sides, we get:

$x\frac{\partial z}{\partial x}=y\frac{\partial z}{\partial y}$

In partial derivative, we take the other variable as a constant.

Concept:

When the input of a function is made up of multiple variables, partial derivative is used to understand how the function changes as we let just one of the variable changes, keeping all the other variables as constant.

Application:

Given $f\left(x,y,z\right)=e^{1-x\cos y}+xz{e}^{-\frac{1}{1+y^2}}$

Differentiating the above with respect to x and treating y and z as a constant, we get:

$\frac{\partial f}{\partial x}=e^{(1-x\cos y)}\cdot \frac{d}{dx}\left(1-x\cos y\right)+z{e}^{\frac{-1}{1+y^2}}\cdot \frac{\partial }{\partial x}\left(x\right)$

$\frac{\partial f}{\partial x}=e^{\left(1-x\cos y\right)}\cdot \left(-\cos y\right)+z{e}^{\frac{-1}{y^2+1}}$

$\frac{\partial f}{\partial x}=-\cos y{e}^{(1-x\cos y)}+z{e}^{-\frac{1}{y^2+1}}$

At point (1, 0, e), the partial derivative gives a value:

$\frac{\partial f}{\partial x}{|}_{\left(1,0,e\right)}=-\cos \left(0\right)e^{(1-\cos 0)}+e.e^{-\frac{1}{1}}$

= -1e⁰ + e.e^-1

= -1 + 1

We are given z = e^{ax + by}⋅ f(ax - by)

Now, $\frac{\partial z}{\partial x}=a\cdot e^{ax+by}\cdot F\left(ax-by\right)+a{e}^{ax+by}\cdot F^{\prime }\left(ax-by\right)$

$b\frac{\partial z}{\partial x}=ab[z+e^{ax+by}\cdot F^{\prime }\left(ax-by\right)$ ---(1)

Now, $\frac{\partial z}{\partial y}=b\cdot e^{ax+by}\cdot F\left(ax-by\right)-b\cdot e^{ax+by}\cdot F^{\prime }\left(ax-by\right)$

$a\frac{\partial z}{\partial y}=ab\left[z-e^{\left(ax+by\right)}\cdot F^{\prime }\left(ax-by\right)\right]$ ---(2)

Add equation (1) & (2), we get

CONCEPT:

In Partial Differentiation, all variables are considered as a constant except the independent derivative variable i.e If f(x,y,z) is a function, then its partial derivative with respect to x is calculated by keeping y and z as constant.

CALCULATION:

Given x² + y² = F (say…)

and 6y + 4x = G (say…)

Differentiating F with respect to y and G with respect to x and then equating them according to the given condition, we get;

2y = 4

So, y = 2.