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

Concept:

The total derivative $\frac{du}{dx}$ is given by:

$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$

$\frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx}$

Calculation:

u = log xy    where x2 + y2 = 1

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

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

x2 + y2 = 1

$\frac{dy}{dx}=\frac{-x}{y}$

$\frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx}$

$\frac{du}{dx}=\frac{1}{x}+\frac{1}{y}(\frac{-x}{y})$

$\frac{du}{dx}=\frac{1}{x}-\frac{x}{y^2}$

Concept -

Newton's Forward Difference Formula:

The Newton's forward difference formula for the first derivative at x = x₀ (where  x₀ is the first value in the table, i.e., x = 1 is given by:

$f^{\prime }(x_0)\approx \frac{1}{h}(\mathrm{\Delta }{f}_0-\frac{1}{2}{\mathrm{\Delta }}^2{f}_0+\frac{1}{3}{\mathrm{\Delta }}^3{f}_0-\cdots )$

where h is the interval between the x-values (here h = 1).

Explanation -

Given the table:

$\begin{array}{cc}x& f(x)\\ 1& 20\\ 2& 22\\ 3& 27\\ 4& 35\end{array}$

Forward Differences:

Let's calculate the forward differences $\mathrm{\Delta }f$:

$\begin{array}{ccccc}x& f(x)& \mathrm{\Delta }f& {\mathrm{\Delta }}^{2}f& {\mathrm{\Delta }}^{3}f\\ 1& 20& 2& 3& 3\\ 2& 22& 5& 6& \\ 3& 27& 8& & \\ 4& 35& & & \end{array}$

For the value f'(4)  using the backward difference approach:

$f^{\prime }(4)\approx \frac{1}{h}(\mathrm{\Delta }{f}_3+\frac{1}{2}{\mathrm{\Delta }}^2{f}_2+\frac{1}{3}{\mathrm{\Delta }}^3{f}_1)$

Step-by-Step Solution:

1. Calculate $\mathrm{\Delta }f$ :

$\mathrm{\Delta }{f}_1=f(2)-f(1)=22-20=2$

$\mathrm{\Delta }{f}_2=f(3)-f(2)=27-22=5$

$\mathrm{\Delta }{f}_3=f(4)-f(3)=35-27=8$

2. Calculate ${\mathrm{\Delta }}^{2}f$ :

${\mathrm{\Delta }}^2{f}_1=\mathrm{\Delta }{f}_2-\mathrm{\Delta }{f}_1=5-2=3$

${\mathrm{\Delta }}^2{f}_2=\mathrm{\Delta }{f}_3-\mathrm{\Delta }{f}_2=8-5=3$

3. Calculate ${\mathrm{\Delta }}^{3}f$ :

${\mathrm{\Delta }}^3{f}_1={\mathrm{\Delta }}^2{f}_2-{\mathrm{\Delta }}^2{f}_1=3-3=0$

Apply the formula for f'(4) & Since h = 1 :

$f^{\prime }(4)\approx \frac{1}{h}(\mathrm{\Delta }{f}_3+\frac{1}{2}{\mathrm{\Delta }}^2{f}_2+\frac{1}{3}{\mathrm{\Delta }}^3{f}_1)$

⇒ $f^{\prime }(4)\approx 1(8+\frac{1}{2}\times 3+\frac{1}{3}\times 0)$

⇒$f^{\prime }(4)\approx 8+1.5+0$

⇒ f'(4) ≈  9.5

Therefore, the value of f'(x) at x = 4 using Newton's forward difference formula is approximately 9.5 .

Concept:

If z = f(x,y)

then the change in z (total differential) is

 $dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

Calculation:

Given:

z = xy
Then the total differential is

$dz=\frac{\partial (xy)}{\partial x}dx+\frac{\partial (xy)}{\partial y}dy$

z = ydx + xdy

Explanation:

Given the function is, $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)={\mathrm{x}}^3\mathrm{y}-\mathrm{x}{\mathrm{y}}^3$ and $\left[\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{x}}}+\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{y}}}\right]$ is to be found out.

Now as the given function is given as function of both x and y, so at first partial derivative is to be

calculated and then the value of x and y is to be substituted to find out absolute derivative.

∴ $\frac{\partial f}{\partial x}=3x^{2}y-y^3\therefore \frac{\partial f}{\partial x}{|}_{x=1,y=2}=3\times 1^2\times 2-2^3=-2$

∴ $\frac{\partial f}{\partial y}=x^3-3y^{2}x\therefore \frac{\partial f}{\partial y}{|}_{x=1,y=2}=1^3-3\times 2^2\times 1=-11$

∴ $Valueof\left[\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{x}}}+\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{y}}}\right]at\mathrm{x}=1,\mathrm{y}=2=\frac{1}{\frac{\partial f}{\partial x}{|}_{x=1,y=2}}+\frac{1}{\frac{\partial f}{\partial y}{|}_{x=1,y=2}}=\frac{1}{-2}+\frac{1}{-11}=-\frac{13}{22}$

Concept:

If W = f(x, y, z) is a continuous function of n variables x, y, z, …, with continuous partial derivatives ∂W/∂x, ∂W/∂y, ∂W/∂z, … and if x, y, z, … are differentiable functions x = x(t), y = y(t), z = z(t), etc. of a variable t, then the total derivative of w with respect to t is given by

$\frac{dW}{dt}=\frac{\partial W}{\partial x}\frac{dx}{dt}+\frac{\partial W}{\partial y}\frac{dy}{dt}+\frac{\partial W}{\partial z}\frac{dz}{dt}$

Calculation:

f(x, y) = e^x sin y

$\frac{\partial f}{\partial x}=e^x\sin y$

$\frac{\partial f}{\partial y}=e^x\cos y$

At t = 0, $\frac{\partial f}{\partial x}=e\sin 0=0$

$\frac{\partial f}{\partial y}=e^1\cos 0=e$

x = t³ + 1 and y = t⁴ + t

$\frac{dx}{dt}=3t^2,\frac{dy}{dt}=4t^3+1$

At t = 0, x = 1 and y = 0

At t = 0, $\frac{dx}{dt}=0,\frac{dy}{dt}=1$

$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$

= 0 + e = e = 2.718

Concept:

If z = f(x,y)

then the change in z (total differential) is

 $dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

Calculation:

Given:

z = xy
Then the total differential is

$dz=\frac{\partial (xy)}{\partial x}dx+\frac{\partial (xy)}{\partial y}dy$

z = ydx + xdy

Explanation:

Given the function is, $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)={\mathrm{x}}^3\mathrm{y}-\mathrm{x}{\mathrm{y}}^3$ and $\left[\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{x}}}+\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{y}}}\right]$ is to be found out.

Now as the given function is given as function of both x and y, so at first partial derivative is to be

calculated and then the value of x and y is to be substituted to find out absolute derivative.

∴ $\frac{\partial f}{\partial x}=3x^{2}y-y^3\therefore \frac{\partial f}{\partial x}{|}_{x=1,y=2}=3\times 1^2\times 2-2^3=-2$

∴ $\frac{\partial f}{\partial y}=x^3-3y^{2}x\therefore \frac{\partial f}{\partial y}{|}_{x=1,y=2}=1^3-3\times 2^2\times 1=-11$

∴ $Valueof\left[\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{x}}}+\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{y}}}\right]at\mathrm{x}=1,\mathrm{y}=2=\frac{1}{\frac{\partial f}{\partial x}{|}_{x=1,y=2}}+\frac{1}{\frac{\partial f}{\partial y}{|}_{x=1,y=2}}=\frac{1}{-2}+\frac{1}{-11}=-\frac{13}{22}$

Concept:

The total derivative $\frac{du}{dx}$ is given by:

$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$

$\frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx}$

Calculation:

u = log xy    where x2 + y2 = 1

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

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

x2 + y2 = 1

$\frac{dy}{dx}=\frac{-x}{y}$

$\frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx}$

$\frac{du}{dx}=\frac{1}{x}+\frac{1}{y}(\frac{-x}{y})$

$\frac{du}{dx}=\frac{1}{x}-\frac{x}{y^2}$

x = a (1 + sin θ)

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\theta }$ = a (cos θ)

y = a cos²θ

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\theta }$ = -2 x a x cos θ × sin θ

$\begin{array}{l}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}/\mathrm{d}\theta }{\mathrm{d}\mathrm{x}/\mathrm{d}\theta }=\frac{-2\mathrm{a}\cos \theta \sin \theta }{\mathrm{a}\cos \theta }\\ \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-2\times \sin \theta \end{array}$

$\begin{array}{l}\sin \theta =\frac{x}{a}-1\\ \frac{dy}{dx}=-2\left(\frac{x}{a}-1\right)\\ \frac{d^{2}y}{d{x}^2}=-\frac{2}{a}\end{array}$

Concept:

If W = f(x, y, z) is a continuous function of n variables x, y, z, …, with continuous partial derivatives ∂W/∂x, ∂W/∂y, ∂W/∂z, … and if x, y, z, … are differentiable functions x = x(t), y = y(t), z = z(t), etc. of a variable t, then the total derivative of w with respect to t is given by

$\frac{dW}{dt}=\frac{\partial W}{\partial x}\frac{dx}{dt}+\frac{\partial W}{\partial y}\frac{dy}{dt}+\frac{\partial W}{\partial z}\frac{dz}{dt}$

Calculation:

f(x, y) = e^x sin y

$\frac{\partial f}{\partial x}=e^x\sin y$

$\frac{\partial f}{\partial y}=e^x\cos y$

At t = 0, $\frac{\partial f}{\partial x}=e\sin 0=0$

$\frac{\partial f}{\partial y}=e^1\cos 0=e$

x = t³ + 1 and y = t⁴ + t

$\frac{dx}{dt}=3t^2,\frac{dy}{dt}=4t^3+1$

At t = 0, x = 1 and y = 0

At t = 0, $\frac{dx}{dt}=0,\frac{dy}{dt}=1$

$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$

= 0 + e = e = 2.718

Concept:

If z = f(x,y)

then the change in z (total differential) is

 $dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

Calculation:

Given:

z = xy
Then the total differential is

$dz=\frac{\partial (xy)}{\partial x}dx+\frac{\partial (xy)}{\partial y}dy$

z = ydx + xdy

$f\left(x\right)=\left|sinx\right|$ is

Option (b) is incorrect f(x) is not differentiable at

$x=0,\pm \pi ,\pm 2\pi ,\pm 3\pi \dots \dots \dots$

→ f(x) is continuous for all x

→ f(3π) = 0

→ for $x>0,\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)$ can be any  value between 0 to 1

Explanation:

Given the function is, $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)={\mathrm{x}}^3\mathrm{y}-\mathrm{x}{\mathrm{y}}^3$ and $\left[\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{x}}}+\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{y}}}\right]$ is to be found out.

Now as the given function is given as function of both x and y, so at first partial derivative is to be

calculated and then the value of x and y is to be substituted to find out absolute derivative.

∴ $\frac{\partial f}{\partial x}=3x^{2}y-y^3\therefore \frac{\partial f}{\partial x}{|}_{x=1,y=2}=3\times 1^2\times 2-2^3=-2$

∴ $\frac{\partial f}{\partial y}=x^3-3y^{2}x\therefore \frac{\partial f}{\partial y}{|}_{x=1,y=2}=1^3-3\times 2^2\times 1=-11$

∴ $Valueof\left[\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{x}}}+\frac{1}{\frac{\mathrm{d}\mathrm{f}}{\mathrm{d}\mathrm{y}}}\right]at\mathrm{x}=1,\mathrm{y}=2=\frac{1}{\frac{\partial f}{\partial x}{|}_{x=1,y=2}}+\frac{1}{\frac{\partial f}{\partial y}{|}_{x=1,y=2}}=\frac{1}{-2}+\frac{1}{-11}=-\frac{13}{22}$

Concept:

The total derivative $\frac{du}{dx}$ is given by:

$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$

$\frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx}$

Calculation:

u = log xy    where x2 + y2 = 1

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

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

x2 + y2 = 1

$\frac{dy}{dx}=\frac{-x}{y}$

$\frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx}$

$\frac{du}{dx}=\frac{1}{x}+\frac{1}{y}(\frac{-x}{y})$

$\frac{du}{dx}=\frac{1}{x}-\frac{x}{y^2}$

Concept -

Newton's Forward Difference Formula:

The Newton's forward difference formula for the first derivative at x = x₀ (where  x₀ is the first value in the table, i.e., x = 1 is given by:

$f^{\prime }(x_0)\approx \frac{1}{h}(\mathrm{\Delta }{f}_0-\frac{1}{2}{\mathrm{\Delta }}^2{f}_0+\frac{1}{3}{\mathrm{\Delta }}^3{f}_0-\cdots )$

where h is the interval between the x-values (here h = 1).

Explanation -

Given the table:

$\begin{array}{cc}x& f(x)\\ 1& 20\\ 2& 22\\ 3& 27\\ 4& 35\end{array}$

Forward Differences:

Let's calculate the forward differences $\mathrm{\Delta }f$:

$\begin{array}{ccccc}x& f(x)& \mathrm{\Delta }f& {\mathrm{\Delta }}^{2}f& {\mathrm{\Delta }}^{3}f\\ 1& 20& 2& 3& 3\\ 2& 22& 5& 6& \\ 3& 27& 8& & \\ 4& 35& & & \end{array}$

For the value f'(4)  using the backward difference approach:

$f^{\prime }(4)\approx \frac{1}{h}(\mathrm{\Delta }{f}_3+\frac{1}{2}{\mathrm{\Delta }}^2{f}_2+\frac{1}{3}{\mathrm{\Delta }}^3{f}_1)$

Step-by-Step Solution:

1. Calculate $\mathrm{\Delta }f$ :

$\mathrm{\Delta }{f}_1=f(2)-f(1)=22-20=2$

$\mathrm{\Delta }{f}_2=f(3)-f(2)=27-22=5$

$\mathrm{\Delta }{f}_3=f(4)-f(3)=35-27=8$

2. Calculate ${\mathrm{\Delta }}^{2}f$ :

${\mathrm{\Delta }}^2{f}_1=\mathrm{\Delta }{f}_2-\mathrm{\Delta }{f}_1=5-2=3$

${\mathrm{\Delta }}^2{f}_2=\mathrm{\Delta }{f}_3-\mathrm{\Delta }{f}_2=8-5=3$

3. Calculate ${\mathrm{\Delta }}^{3}f$ :

${\mathrm{\Delta }}^3{f}_1={\mathrm{\Delta }}^2{f}_2-{\mathrm{\Delta }}^2{f}_1=3-3=0$

Apply the formula for f'(4) & Since h = 1 :

$f^{\prime }(4)\approx \frac{1}{h}(\mathrm{\Delta }{f}_3+\frac{1}{2}{\mathrm{\Delta }}^2{f}_2+\frac{1}{3}{\mathrm{\Delta }}^3{f}_1)$

⇒ $f^{\prime }(4)\approx 1(8+\frac{1}{2}\times 3+\frac{1}{3}\times 0)$

⇒$f^{\prime }(4)\approx 8+1.5+0$

⇒ f'(4) ≈  9.5

Therefore, the value of f'(x) at x = 4 using Newton's forward difference formula is approximately 9.5 .