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:

\(\rm\frac{X}{(X+2)(X−3)}= \rm\frac{A}{(X+2)}+\rm\frac{B}{(X−3)} ~~~~~---- Eq(i) \\ \therefore \rm\frac{X}{(X+2)(X−3)}= \rm\frac{A(x-3)+B(x+2)}{(x +2)(X-3)} \\ \Rightarrow \rm\frac{X}{(X+2)(X−3)}= \rm\frac{x(A+B)+2B-3A}{(x +2)(X-3)} \)

Now comparing both sides:

A + B = 1  ----- ---Eq(ii)

-3A + 2B = 0  ----Eq(iiI)

after solving the above equations, we get:

\(\rm{A = \frac{2}{5}\ \& ~ B = \frac{3}{5}} \)

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

\(\rm\frac{X}{(X+2)(X−3)} = \rm\frac{2}{5(X+2)}+\frac{3}{5(X−3)}\)

\(\rm \frac{\partial u}{\partial z} = xye^{xyz}\)

\(\rm \frac{\partial }{\partial y} \left( \rm \frac{\partial u}{\partial z} \right)= \rm \frac{\partial }{\partial y}(xye^{xyz})\)

\(\rm \frac{\partial^2 u}{\partial y \partial z} = xy \rm \frac{\partial }{\partial y}(e^{xyz}) + e^{xyz} \rm \frac{\partial }{\partial y}(xy)\)

= xy(xz)e^xyz + xe^xyz

\(\rm \frac{\partial }{\partial x} \left(\frac{\partial^2 u}{\partial y \partial z} \right) = \rm \frac{\partial }{\partial x}(x^2 yz + x) e^{xyz}\)

\(\rm \frac{\partial^3 u}{\partial x \partial y \partial z} = (x^2 yz + x) yze^{xyz} + e^{xyz} (2xyz + 1)\)

\(\rm = e^{xyz} (x^2 y^2 z^2 + xyz + 2xyz + 1)\)

= (1 + 3xyz + x²y²z²) e^xyz

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

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

= 5e

Concept:

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

\(\frac{{\partial \left( {{y_1},{y_2},\;{y_3}} \right)}}{{\partial \left( {{x_1},\;{x_2},\;{x_3}} \right)}} = \left| {\begin{array}{*{20}{c}} {\frac{{\partial {y_1}}}{{\partial {x_1}}}}&{\frac{{\partial {y_1}}}{{\partial {x_2}}}}&{\frac{{\partial {y_1}}}{{\partial {x_3}}}}\\ {\frac{{\partial {y_2}}}{{\partial {x_1}}}}&{\frac{{\partial {y_2}}}{{\partial {x_2}}}}&{\frac{{\partial {y_2}}}{{\partial {x_3}}}}\\ {\frac{{\partial {y_3}}}{{\partial {x_1}}}}&{\frac{{\partial {y_3}}}{{\partial {x_2}}}}&{\frac{{\partial {y_3}}}{{\partial {x_3}}}} \end{array}} \right|\)

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

Calculation:

Given:

x = uv and \({\rm{y}} = \frac{{{\rm{u}} + {\rm{v}}}}{{{\rm{u}} - {\rm{v}}}}\)

\(\frac{{\partial \left( {{\rm{x}},{\rm{y}}} \right)}}{{\partial \left( {{\rm{u}},{\rm{v}}} \right)}} = \left| {\begin{array}{*{20}{c}} {\frac{{\partial {\rm{x}}}}{{\partial {\rm{u}}}}}&{\frac{{\partial {\rm{x}}}}{{\partial {\rm{v}}}}}\\ {\frac{{\partial {\rm{y}}}}{{\partial {\rm{u}}}}}&{\frac{{\partial {\rm{y}}}}{{\partial {\rm{v}}}}} \end{array}} \right| \)

\(\frac{{\partial \left( {{\rm{x}},{\rm{y}}} \right)}}{{\partial \left( {{\rm{u}},{\rm{v}}} \right)}} =\left| {\begin{array}{*{20}{c}} {\rm{v}}&{\rm{u}}\\ { - \frac{{2{\rm{v}}}}{{{{\left( {{\rm{u}} - {\rm{v}}} \right)}^2}}}}&{\frac{{2{\rm{u}}}}{{{{\left( {{\rm{u}} - {\rm{v}}} \right)}^2}}}} \end{array}} \right|\)

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

We know that

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

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

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

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

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 

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

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

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

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

\(x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} = nf\)

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

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

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

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:

\(\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\)

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

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

Calculation:

Given: y = log sinx

Let sin x = u

⇒ y = log u

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

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

Hence, the value of \(\frac{dy}{dx}\) will be \(\frac{1}{sin~x} cos~x\).

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

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

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

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

Substitute in (1)

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

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:

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

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

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

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

Similarly,

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

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

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

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

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

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

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

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

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:

\(\frac{{{\rm{dw}}}}{{{\rm{dt}}}} = \frac{{\partial {\rm{w}}}}{{\partial {\rm{x}}}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}} + \frac{{\partial {\rm{w}}}}{{\partial {\rm{y}}}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}}\)

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

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

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:

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

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 1\)

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 

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

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

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

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

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

\(\frac {\partial f}{\partial y} = - \frac {\partial g}{\partial x}\)

\(f = {y^x}\)

ln f = x lny

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

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

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

\( = {y^{x - 1}} + x{y^{x - 1}}lny\)

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