Partial Derivatives: 15 MCQs with answers

The rate of change of $f$ with respect to $x$, holding all other variables constant

The partial derivative with respect to $x$ measures how $f$ changes as $x$ changes, while keeping all other independent variables fixed.

$x^3 y^2$

Differentiate with respect to $z$, treating $x$ and $y$ as constants. The derivative of $z$ is 1, so the result is $x^3 y^2$.

$2x + 2y$

$\frac{\partial f}{\partial x} = 2x$ and $\frac{\partial f}{\partial y} = 2y$. Their sum is $2x + 2y$.

$3x^2 y^2$

Differentiate with respect to $y$, treating $x$ as a constant: $\frac{\partial}{\partial y}(x^2 y^3) = x^2 \cdot 3y^2 = 3x^2 y^2$.

$\frac{1}{y}$

Treat $y$ as a constant. The derivative of $x$ with respect to $x$ is 1, so $\frac{\partial}{\partial x} \left(\frac{x}{y}\right) = \frac{1}{y}$.

$3x + 2y$

Differentiate with respect to $y$: $\frac{\partial}{\partial y}(x^2) = 0$, $\frac{\partial}{\partial y}(3xy) = 3x$, $\frac{\partial}{\partial y}(y^2) = 2y$. So the result is $3x + 2y$.

True

This is the definition of the partial derivative with respect to $x$: it is the ordinary derivative of $f$ as a function of $x$ alone, with $y$ held constant.

$2xy + 3y^3$

Differentiate with respect to $x$ treating $y$ as a constant: $\frac{\partial}{\partial x}(x^2 y) = 2xy$ and $\frac{\partial}{\partial x}(3xy^3) = 3y^3$. So the result is $2xy + 3y^3$.

$x e^{xy} \sin x$

Treat $x$ as a constant. The derivative of $e^{xy}$ with respect to $y$ is $x e^{xy}$, and $\sin x$ is a constant factor. So $\frac{\partial f}{\partial y} = x e^{xy} \sin x$.

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

Using the chain rule: $\frac{\partial}{\partial x} \ln(x^2 + y^2) = \frac{1}{x^2 + y^2} \cdot 2x = \frac{2x}{x^2 + y^2}$.

$\frac{y}{\sqrt{x^2 + y^2}}$

Let $u = x^2 + y^2$. Then $f = \sqrt{u}$, so $\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{u}} \cdot 2y = \frac{y}{\sqrt{x^2 + y^2}}$.

$y \cos(xy)$

Using the chain rule: $\frac{\partial}{\partial x} \sin(xy) = \cos(xy) \cdot y = y \cos(xy)$.

False

They are equal only if the mixed partial derivatives are continuous (Clairaut's theorem). In general, they may differ.

$2x$

First compute $\frac{\partial f}{\partial y} = x^2 + 3y^2$. Then differentiate with respect to $x$: $\frac{\partial}{\partial x}(x^2 + 3y^2) = 2x$.

$6x^2 y$

First compute $\frac{\partial f}{\partial x} = 3x^2 y^2$. Then differentiate with respect to $y$: $\frac{\partial}{\partial y}(3x^2 y^2) = 6x^2 y$.