Partial Derivatives MCQ Quiz - Objective Question with Answer for Partial Derivatives - Download Free PDF

Explanation:

x = x2 - x y + y3,

x = rcosθ, y = rsinθ

Apply Chain Rule:

(dz/dr) = (dz/dx) (dx/dr) + (dz/dy) (dy/dr)

dz/dx = 2x - y

dz/dy = -x + 3y²

dx/dr = cos(θ)

dy/dr = sin(θ)

Now,  (dz/dr) = (2x - y) cos(θ) + (-x + 3y²) sin(θ)

Since x = r cos(θ) and y = r sin(θ), we can find r and θ at x = 1 and y = 1:

r = √(x² + y²) = √(1² + 1²) = √2

θ = tan⁻¹(y/x) = tan⁻¹(1/1) = π/4

Now, substitute x = 1, y = 1, r = √2, and θ = π/4 into the expression for (dz/dr):

(dz/dr) = (2(1) - 1) cos(π/4) + (-1 + 3(1)²) sin(π/4)

(dz/dr) = 1 (√2/2) + 2 (√2/2)

(dz/dr) = 3√2/2

Therefore, (dz/dr) at x = 1 and y = 1 is 3/√2 

Hence, the correct answer is option 1.

Let's solve it using partial diffraction:

$$\begin{gathered} \frac{\mathrm{X}}{(\mathrm{X}+2)(\mathrm{X}-3)}=\frac{\mathrm{A}}{(\mathrm{X}+2)}+\frac{\mathrm{B}}{(\mathrm{X}-3)}----\mathrm{E}\mathrm{q}(\mathrm{i}) \\ \therefore \frac{\mathrm{X}}{(\mathrm{X}+2)(\mathrm{X}-3)}=\frac{\mathrm{A}(\mathrm{x}-3)+\mathrm{B}(\mathrm{x}+2)}{(\mathrm{x}+2)(\mathrm{X}-3)} \\ \Rightarrow \frac{\mathrm{X}}{(\mathrm{X}+2)(\mathrm{X}-3)}=\frac{\mathrm{x}(\mathrm{A}+\mathrm{B})+2\mathrm{B}-3\mathrm{A}}{(\mathrm{x}+2)(\mathrm{X}-3)} \end{gathered}$$

Now comparing both sides:

A + B = 1  ----- ---Eq(ii)

-3A + 2B = 0  ----Eq(iiI)

after solving the above equations, we get:

$\mathrm{A}=\frac{2}{5}\mathrm{\&}\mathrm{B}=\frac{3}{5}$

after putting these value in Eq(i), we will get:

$\frac{\mathrm{X}}{(\mathrm{X}+2)(\mathrm{X}-3)}=\frac{2}{5(\mathrm{X}+2)}+\frac{3}{5(\mathrm{X}-3)}$

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

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

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

= 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:

Jacobian of (y1, y2, y3) w.r.t. (x1, x2, x3) is equal to

$\frac{\partial \left(y_1,y_2,y_3\right)}{\partial \left(x_1,x_2,x_3\right)}=\left|\begin{array}{ccc}\frac{\partial y_1}{\partial x_1}& \frac{\partial y_1}{\partial x_2}& \frac{\partial y_1}{\partial x_3}\\ \frac{\partial y_2}{\partial x_1}& \frac{\partial y_2}{\partial x_2}& \frac{\partial y_2}{\partial x_3}\\ \frac{\partial y_3}{\partial x_1}& \frac{\partial y_3}{\partial x_2}& \frac{\partial y_3}{\partial x_3}\end{array}\right|$

$\frac{\partial \left(x_1,x_2,x_3\right)}{\partial \left(y_1,y_2,y_3\right)}=\frac{1}{\frac{\partial \left(y_1,y_2,y_3\right)}{\partial \left(x_1,x_2,x_3\right)}}$

Calculation:

Given:

x = uv and $\mathrm{y}=\frac{\mathrm{u}+\mathrm{v}}{\mathrm{u}-\mathrm{v}}$

$\frac{\partial \left(\mathrm{x},\mathrm{y}\right)}{\partial \left(\mathrm{u},\mathrm{v}\right)}=\left|\begin{array}{cc}\frac{\partial \mathrm{x}}{\partial \mathrm{u}}& \frac{\partial \mathrm{x}}{\partial \mathrm{v}}\\ \frac{\partial \mathrm{y}}{\partial \mathrm{u}}& \frac{\partial \mathrm{y}}{\partial \mathrm{v}}\end{array}\right|$

$\frac{\partial \left(\mathrm{x},\mathrm{y}\right)}{\partial \left(\mathrm{u},\mathrm{v}\right)}=\left|\begin{array}{cc}\mathrm{v}& \mathrm{u}\\ -\frac{2\mathrm{v}}{{\left(\mathrm{u}-\mathrm{v}\right)}^2}& \frac{2\mathrm{u}}{{\left(\mathrm{u}-\mathrm{v}\right)}^2}\end{array}\right|$

$\frac{\partial \left(\mathrm{x},\mathrm{y}\right)}{\partial \left(\mathrm{u},\mathrm{v}\right)}=\frac{4\mathrm{u}\mathrm{v}}{{\left(\mathrm{u}-\mathrm{v}\right)}^2}$

We know that

$\frac{\partial \left(x_1,x_2,x_3\right)}{\partial \left(y_1,y_2,y_3\right)}=\frac{1}{\frac{\partial \left(y_1,y_2,y_3\right)}{\partial \left(x_1,x_2,x_3\right)}}$

$\frac{\partial \left(\mathrm{u},\mathrm{v}\right)}{\partial \left(\mathrm{x},\mathrm{y}\right)}=\frac{1}{\frac{\partial \left(\mathrm{x},\mathrm{y}\right)}{\partial \left(\mathrm{u},\mathrm{v}\right)}}$

$\therefore \frac{\partial \left(\mathbf{u},\mathbf{v}\right)}{\partial \left(\mathbf{x},\mathbf{y}\right)}=\frac{{\left(\mathbf{u}-\mathbf{v}\right)}^2}{4\mathbf{u}\mathbf{v}}=f(u,v)$

$f(2,1)=\frac{{\left(\mathbf{2}-\mathbf{1}\right)}^2}{4\times 2\times 1}=\frac{1}{8}=0.125$

Concept:

Partial derivative: A partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant.

Calculation:

Given:

f (x, y) = x3 – 3xy2 and g(x,y) = 3x2y – y3 

$\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}(x^3-3x{y}^2)=$ 3x² - 3y²

$\frac{\partial g}{\partial x}=\frac{\partial }{\partial x}(3x^{2}y-y^3)=$ 6xy

$\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}(x^3-3x{y}^2)=$ -6xy

$\frac{\partial g}{\partial y}=\frac{\partial }{\partial y}(3x^{2}y-y^3)=$ 3x² - 3y²

From the above values, only option (3) is correct i.e.

$\frac{\partial f}{\partial y}=-\frac{\partial g}{\partial x}$

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

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

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

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, $u=\ln \left(\frac{x^3+x^{2}y-y^3}{x-y}\right)$

$z=\frac{x^3+x^{2}y-y^3}{x-y}$

z is a homogenous function of x & y with a degree 2.

Now, z = eu

Thus, by Euler’s theorem:

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2\frac{e^u}{e^u}$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2$

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:

f(x, y, z) = (x² + y² – 2z²)(y² + z²)

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

At the point, x = 2, y = 1 and z = 3 is

$\frac{\partial f}{\partial x}=2\left(2\right)\left(1^2+3^2\right)=40$

Concept:

Chain Rule of derivatives states that, if y = f(u) and u = g(x) are both differentiable functions, then:

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

$\frac{d\left(\ln x\right)}{dx}=\frac{1}{x},forx>0$

$\frac{d\left(\sin x\right)}{dx}=cosx$

Calculation:

Given: y = log sinx

Let sin x = u

⇒ y = log u

$\frac{d}{dx}\left(\log u\right)=\frac{1}{u}\frac{d}{dx}\left(u\right)$

$=\frac{1}{\sin x}\left(cosx\right)$

Hence, the value of $\frac{dy}{dx}$ will be $\frac{1}{sinx}cosx$.

r = x2 + y - z   ---(1)

z3 -xy + yz + y3 = 1    ---(2)

$\frac{\partial r}{\partial x}=2x+\frac{\partial y}{\partial x}-\frac{\partial z}{\partial x}$

Since y is an independent derivative of ‘y’ w.r.t. ‘x’ is 0.

$\frac{\partial r}{\partial x}=2x-\frac{\partial z}{\partial x}$      ----(1)

From 2^nd relation:

Z³ – xy + yz + y³ = 1

Differentiate w.r.t x

$3Z^2\frac{\partial z}{\partial x}-y+y\frac{\partial z}{\partial x}=0$

$\left(3Z^2+y\right)\frac{\partial z}{\partial x}=y$

$\frac{\partial z}{\partial x}=\frac{y}{3z^2+y}$      ----(2)

Substitute in (1)

$\frac{\partial r}{\partial x}=2r-\frac{y}{3z^2+y}$

At, (2, -1, 1)

${\left(\frac{\partial r}{\partial x}\right)}_{\left(2,-1,1\right)}=2\left(2\right)-\frac{-1}{3{\left(1\right)}^2+\left(-1\right)}$

$=4+\frac{1}{2}$

Concept

if y = xⁿ, then$\frac{\partial y}{\partial x}=n{X}^{n-1}\frac{{\partial }^{2}y}{\partial x^2}=n\left(n-1\right)X^{n-2}$

Calculation:

Given:

$v={\left(x^2+y^2+z^2\right)}^{-\frac{1}{2}}$

$\frac{\partial v}{\partial x}=-\frac{1}{2}{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\times 2x=-x{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}$

$\frac{{\partial }^{2}v}{\partial x^2}=-x\left\{-\frac{3}{2}{\left(x^2+y^2+z^2\right)}^{-\frac{5}{2}}\times 2x\right\}+{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}\left(-1\right)$

$\Rightarrow 3x^2{\left(x^2+y^2+z^2\right)}^{-\frac{5}{2}}-{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}$

Similarly,

$\frac{{\partial }^{2}v}{\partial y^2}=3y^2{\left(x^2+y^2+z^2\right)}^{-\frac{5}{2}}-{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}$

$\frac{{\partial }^{2}v}{\partial z^2}=3z^2{\left(x^2+y^2+z^2\right)}^{-\frac{5}{2}}-{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}$

Now,$\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}v}{\partial z^2}=\left(3x^2+3y^2+3z^2\right){\left(x^2+y^2+z^2\right)}^{-\frac{5}{2}}-3{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}$

⇒$3\left(x^2+y^2+z^2\right){\left(x^2+y^2+z^2\right)}^{-\frac{5}{2}}-3{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}$

⇒$3{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}-3{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}=0$

$f\left(x,y\right)=\frac{a{x}^2+b{y}^2}{xy}=\frac{ax}{y}+\frac{by}{x}$

$\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left[\frac{ax}{y}+\frac{by}{x}\right]$

$\frac{\partial f}{\partial x}=\frac{a}{y}-\frac{by}{x^2}$

At, x = 1 and y = 2, we get:

$\frac{\partial f}{\partial x}{|}_{x=1,y=2}=\frac{a}{2}-\frac{b\left(2\right)}{1^2}$

$=\frac{a}{2}-2b$   ----(1)

Similarly,

$\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left[\frac{ax}{y}+\frac{by}{x}\right]$

$=-\frac{ax}{y^2}+\frac{b}{x}$

At, x = 1 and y = 2

$\frac{\partial f}{\partial y}{|}_{x=1,y=2}=\frac{-a}{4}+b$     ----(2)

For the given condition, equating the value of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at x = 1 and y = 2, we get:

$\frac{a}{2}-2b=\frac{-a}{4}+b$

$\frac{a}{2}+\frac{a}{4}=3b$

$\Rightarrow \frac{2a+a}{4}=3b$

3a = 12b

w = f (x, y)

Partial differentiation with respect to x and y gives:

$\frac{\mathrm{d}\mathrm{w}}{\mathrm{d}\mathrm{t}}=\frac{\partial \mathrm{w}}{\partial \mathrm{x}}\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}+\frac{\partial \mathrm{w}}{\partial \mathrm{y}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}$

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, $u=log\left(\frac{x^2+y^2}{x+y}\right)$

$z=\frac{x^2+y^2}{x+y}$

z is a homogenous function of x & y with a degree 1.

Now, z = eu

Thus, by Euler’s theorem:

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

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=1$

$f=y^x$

ln f = x lny

⇒ $\frac{1}{f}\frac{df}{dy}=\frac{x}{y}$

⇒ $\frac{\partial f}{\partial y}=y^x\left(\frac{x}{y}\right)=y^{x-1}.x$

⇒ $\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}\left(y^{x-1}.x\right)$

$=y^{x-1}+x{y}^{x-1}lny$

$=1^{\left(2-1\right)}+\left[2\times 1^{\left(2-1\right)}\ln \left(1\right)\right]=1$

Concept:

Partial derivative: A partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant.

Calculation:

Given:

f (x, y) = x3 – 3xy2 and g(x,y) = 3x2y – y3 

$\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}(x^3-3x{y}^2)=$ 3x² - 3y²

$\frac{\partial g}{\partial x}=\frac{\partial }{\partial x}(3x^{2}y-y^3)=$ 6xy

$\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}(x^3-3x{y}^2)=$ -6xy

$\frac{\partial g}{\partial y}=\frac{\partial }{\partial y}(3x^{2}y-y^3)=$ 3x² - 3y²

From the above values, only option (3) is correct i.e.

$\frac{\partial f}{\partial y}=-\frac{\partial g}{\partial x}$