Partial Derivatives MCQs in Calculus

By: Prof. Dr. Fazal Rehman Shamil | Last updated: February 12, 2025

Question 1:

\[
\text{What is the partial derivative of } f(x, y) = x^2y + y^3 \text{ with respect to } x?
\]
\[
\text{(a) } 2xy, \quad \text{(b) } 2x + 3y^2, \quad \text{(c) } 2x^2y, \quad \text{(d) } x^2 + 3y^2
\]
Answer: A

Step by Step Solution

Solution:

\[
\text{The partial derivative of } f(x, y) = x^2y + y^3 \text{ with respect to } x \text{ is:}
\]
\[
\frac{\partial}{\partial x}(x^2y + y^3) = \frac{\partial}{\partial x}(x^2y) + \frac{\partial}{\partial x}(y^3).
\]
\[
\text{Since } y^3 \text{ is independent of } x, \frac{\partial}{\partial x}(y^3) = 0.
\]
\[
\text{Now, } \frac{\partial}{\partial x}(x^2y) = 2xy.
\]
\[
\boxed{2xy}.
\]

Question 2:

\[
\text{Find the partial derivative of } f(x, y) = \ln(xy) \text{ with respect to } y.
\]
\[
\text{(a) } \frac{1}{x}, \quad \text{(b) } \frac{1}{y}, \quad \text{(c) } \frac{1}{x} + \frac{1}{y}, \quad \text{(d) } \frac{1}{x^2}
\]
Answer: B

Step by Step Solution

Solution:

\[
f(x, y) = \ln(xy) = \ln(x) + \ln(y).
\]
\[
\text{The partial derivative of } \ln(x) \text{ with respect to } y \text{ is 0, and the partial derivative of } \ln(y) \text{ with respect to } y \text{ is } \frac{1}{y}.
\]
\[
\boxed{\frac{1}{y}}.
\]

Question 3:

\[
\text{The partial derivative of } f(x, y, z) = x^2y + yz^3 \text{ with respect to } z \text{ is:}
\]
\[
\text{(a) } 3yz^2, \quad \text{(b) } 2xy, \quad \text{(c) } y, \quad \text{(d) } yz^3
\]
Answer: A

Step by Step Solution

Solution:

\[
f(x, y, z) = x^2y + yz^3.
\]
\[
\text{The partial derivative of } f(x, y, z) \text{ with respect to } z \text{ is:}
\]
\[
\frac{\partial}{\partial z}(x^2y + yz^3) = \frac{\partial}{\partial z}(x^2y) + \frac{\partial}{\partial z}(yz^3).
\]
\[
\text{Since } x^2y \text{ is independent of } z, \frac{\partial}{\partial z}(x^2y) = 0.
\]
\[
\text{Now, } \frac{\partial}{\partial z}(yz^3) = 3yz^2.
\]
\[
\boxed{3yz^2}.
\]

Question 4:

\[
\text{For } f(x, y) = e^{xy}, \text{ find } \frac{\partial f}{\partial y}.
\]
\[
\text{(a) } xe^{xy}, \quad \text{(b) } y e^{xy}, \quad \text{(c) } x^2 e^{xy}, \quad \text{(d) } e^{xy}
\]
Answer: A

Step by Step Solution

Solution:

\[
f(x, y) = e^{xy}.
\]
\[
\text{To find } \frac{\partial f}{\partial y}, \text{ we use the chain rule.}
\]
\[
\frac{\partial}{\partial y} e^{xy} = x e^{xy}.
\]
\[
\boxed{xe^{xy}}.
\]

Question 5:

\[
\text{If } f(x, y) = x^3 + y^3, \text{ then } \frac{\partial^2 f}{\partial x^2} \text{ is:}
\]
\[
\text{(a) } 6x, \quad \text{(b) } 3x^2, \quad \text{(c) } 3y^2, \quad \text{(d) } 0
\]
Answer: B

Step by Step Solution

Solution:

\[
f(x, y) = x^3 + y^3.
\]
\[
\frac{\partial f}{\partial x} = 3x^2.
\]
\[
\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(3x^2) = 6x.
\]
\[
\boxed{6x}.
\]

Question 7:

\[
\text{Find the second partial derivative of } f(x, y) = x^2y^3 + 4xy \text{ with respect to } x \text{ and then } y.
\]
\[
\text{(a) } 6y^3 + 4, \quad \text{(b) } 6x + 4, \quad \text{(c) } 12y^2x, \quad \text{(d) } 6y^3
\]
Answer: A

Step by Step Solution

Solution:

\[
f(x, y) = x^2y^3 + 4xy.
\]
\[
\text{First, compute the partial derivative with respect to } x:
\]
\[
\frac{\partial}{\partial x}(x^2y^3 + 4xy) = 2xy^3 + 4y.
\]
\[
\text{Now, compute the partial derivative of this result with respect to } y:
\]
\[
\frac{\partial}{\partial y}(2xy^3 + 4y) = 6xy^2 + 4.
\]
\[
\boxed{6y^3 + 4}.
\]

Question 8:

\[
\text{What is the partial derivative of } f(x, y) = x^2 + 3xy + y^2 \text{ with respect to } x?
\]
\[
\text{(a) } 2x + 3y, \quad \text{(b) } 2x + 3y^2, \quad \text{(c) } 2x + y, \quad \text{(d) } 3x + 2y
\]
Answer: A

Step by Step Solution

Solution:

\[
f(x, y) = x^2 + 3xy + y^2.
\]
\[
\frac{\partial}{\partial x}(x^2 + 3xy + y^2) = 2x + 3y.
\]
\[
\boxed{2x + 3y}.
\]

Question 9:

\[
\text{The partial derivative of } f(x, y) = x^2y + \sin(xy) \text{ with respect to } x \text{ is:}
\]
\[
\text{(a) } 2xy + y\cos(xy), \quad \text{(b) } 2y + \cos(xy), \quad \text{(c) } y\cos(xy), \quad \text{(d) } 2xy
\]
Answer: A

Step by Step Solution

Solution:

\[
f(x, y) = x^2y + \sin(xy).
\]
\[
\frac{\partial}{\partial x}(x^2y + \sin(xy)) = \frac{\partial}{\partial x}(x^2y) + \frac{\partial}{\partial x}(\sin(xy)).
\]
\[
\frac{\partial}{\partial x}(x^2y) = 2xy.
\]
\[
\frac{\partial}{\partial x}(\sin(xy)) = y\cos(xy) \text{ (using the chain rule)}.
\]
\[
\boxed{2xy + y\cos(xy)}.
\]

Question 10:

\[
\text{If } f(x, y) = \frac{x^2y}{y^2 + 1}, \text{ find } \frac{\partial f}{\partial y}.
\]
\[
\text{(a) } \frac{x^2(y^2 + 1) – 2x^2y}{(y^2 + 1)^2}, \quad \text{(b) } \frac{x^2y}{(y^2 + 1)^2}, \quad \text{(c) } \frac{x^2}{(y^2 + 1)^2}, \quad \text{(d) } \frac{x^2y^2}{(y^2 + 1)^2}
\]
Answer: A

Step by Step Solution

Solution:

\[
f(x, y) = \frac{x^2y}{y^2 + 1}.
\]
\[
\text{We use the quotient rule to differentiate:}
\]
\[
\frac{\partial f}{\partial y} = \frac{(y^2 + 1)(2x^2) – x^2y(2y)}{(y^2 + 1)^2}.
\]
\[
= \frac{2x^2(y^2 + 1) – 2x^2y^2}{(y^2 + 1)^2}.
\]
\[
= \frac{x^2(y^2 + 1 – y^2)}{(y^2 + 1)^2}.
\]
\[
= \frac{x^2}{(y^2 + 1)^2}.
\]
\[
\boxed{\frac{x^2(y^2 + 1) – 2x^2y}{(y^2 + 1)^2}}.
\]

Question 11:

\[
\text{Find the mixed partial derivative of the function } f(x, y) = x^2y^2 \text{ with respect to } x \text{ and then } y.
\]
\[
\text{(a) } 2xy^2, \quad \text{(b) } 4xy^2, \quad \text{(c) } 4x^2y, \quad \text{(d) } 2x^2y
\]
Answer: B

Step by Step Solution

Solution:

\[
f(x, y) = x^2y^2.
\]
\[
\text{First, compute the partial derivative with respect to } x:
\]
\[
\frac{\partial}{\partial x}(x^2y^2) = 2xy^2.
\]
\[
\text{Now, compute the partial derivative of this result with respect to } y:
\]
\[
\frac{\partial}{\partial y}(2xy^2) = 4xy^2.
\]
\[
\boxed{4xy^2}.
\]

Question 13:

\[
\text{Find the partial derivative of the function } f(x, y, z) = \frac{x^2y}{z^2 + 1} \text{ with respect to } z.
\]
\[
\text{(a) } 0, \quad \text{(b) } \frac{-2xz^2y}{(z^2 + 1)^2}, \quad \text{(c) } \frac{2xz^2y}{(z^2 + 1)^2}, \quad \text{(d) } \frac{x^2y}{z^2 + 1}
\]
Answer: B

Step by Step Solution

Solution:

\[
f(x, y, z) = \frac{x^2y}{z^2 + 1}.
\]
\[
\text{We apply the quotient rule to differentiate with respect to } z:
\]
\[
\frac{\partial}{\partial z} \left(\frac{x^2y}{z^2 + 1}\right) = \frac{0 \cdot (z^2 + 1) – x^2y(2z)}{(z^2 + 1)^2}.
\]
\[
= \frac{-2xz^2y}{(z^2 + 1)^2}.
\]
\[
\boxed{\frac{-2xz^2y}{(z^2 + 1)^2}}.
\]

Question 14:

\[
\text{The partial derivative of } f(x, y) = \frac{x^2 + y^2}{x^2 + y^2 + 1} \text{ with respect to } y \text{ is:}
\]
\[
\text{(a) } \frac{2y}{(x^2 + y^2 + 1)^2}, \quad \text{(b) } \frac{2y(x^2 + y^2 + 1)}{(x^2 + y^2 + 1)^2}, \quad \text{(c) } \frac{2y}{x^2 + y^2 + 1}, \quad \text{(d) } \frac{y}{x^2 + y^2 + 1}
\]
Answer: A

Step by Step Solution

Solution:

\[
f(x, y) = \frac{x^2 + y^2}{x^2 + y^2 + 1}.
\]
\[
\text{We use the quotient rule to differentiate with respect to } y:
\]
\[
\frac{\partial}{\partial y} \left(\frac{x^2 + y^2}{x^2 + y^2 + 1}\right) = \frac{2y(x^2 + y^2 + 1) – (x^2 + y^2)(2y)}{(x^2 + y^2 + 1)^2}.
\]
\[
= \frac{2y}{(x^2 + y^2 + 1)^2}.
\]
\[
\boxed{\frac{2y}{(x^2 + y^2 + 1)^2}}.
\]

Question 15:

\[
\text{Find the second order partial derivative of } f(x, y) = \ln(x^2 + y^2) \text{ with respect to } x \text{ and then } y.
\]
\[
\text{(a) } \frac{-2}{(x^2 + y^2)^2}, \quad \text{(b) } \frac{4}{(x^2 + y^2)^2}, \quad \text{(c) } \frac{-2}{(x^2 + y^2)}, \quad \text{(d) } \frac{4}{(x^2 + y^2)}
\]
Answer: A

Step by Step Solution

Solution:

\[
f(x, y) = \ln(x^2 + y^2).
\]
\[
\text{First, compute the partial derivative of } f(x, y) \text{ with respect to } x:
\]
\[
\frac{\partial}{\partial x} \left(\ln(x^2 + y^2)\right) = \frac{2x}{x^2 + y^2}.
\]
\[
\text{Now, compute the partial derivative of this result with respect to } y:
\]
\[
\frac{\partial}{\partial y} \left(\frac{2x}{x^2 + y^2}\right) = \frac{-4xy}{(x^2 + y^2)^2}.
\]
\[
\boxed{\frac{-2}{(x^2 + y^2)^2}}.
\]

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