Partial Derivatives Examples 1

# Partial Derivatives Examples 1

We will now look at some more examples of computing partial derivatives for functions of several variables. Be sure to review the Partial Derivatives page before hand! More examples can be found on the Partial Derivatives Examples 2 page.

Before we look at the following examples, it will be important to recall the following differentiation rules for single variable real-valued functions:

• Rule for Power Functions: $\frac{d}{dx} x^n = nx^{n-1}$.
• Rules for Trigonometric Functions: $\frac{d}{dx} \sin x = \cos x$, $\frac{d}{dx} \cos x = - \sin x$, and $\frac{d}{Dx} \tan x = \sec ^2 x$.
• Rules for Exponential Functions: $\frac{d}{dx} a^x = a^x \ln (a)$ and $\frac{d}{dx} e^x = e^x$.
• Rules for Logarithmic Functions: $\frac{d}{dx} \log_a (x) = \frac{1}{x \ln a}$ and $\frac{d}{dx} \ln x = \frac{1}{x}$.
• The Chain Rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x)$.
• The Product Rule: $\frac{d}{dx} (f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$.
• The Quotient Rule: $\frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right ) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$ provided that $g(x) \neq 0$.

## Example 1

Let $f(x,y) = x^3 \cos (2xy) - xy^2 e^y$. Find $\frac{\partial}{\partial x} f(x,y)$ and $\frac{\partial}{\partial y} f(x,y)$.

To calculate $\frac{\partial}{\partial x} f(x,y)$ we will need to apply the product rule to the first term, and so:

(1)
\begin{align} \quad \frac{\partial}{\partial x} f(x,y) = 3x^2 \cos (2xy) - 2x^3y \sin (2xy) - y^2 e^y \end{align}

To calculate $\frac{\partial}{\partial y} f(x,y)$ we will also need to apply the product rule, this time to the second term of the function, and thus:

(2)
\begin{align} \quad \frac{\partial}{\partial x} f(x,y) = -2x^4 \sin (2xy) - x \left [2ye^y + y^2e^y \right ] \end{align}

## Example 2

Let $f(x, y) = (1 + xy)^{6}$. Find $\frac{\partial}{\partial x} f(x,y)$ and $\frac{\partial}{\partial y} f(x,y)$.

To calculate $\frac{\partial}{\partial x} f(x,y)$ we will need to apply the chain rule, and so:

(3)
\begin{align} \quad \frac{\partial}{\partial x} f(x,y) = 6(1 + xy)^5 \cdot y \end{align}

To calculate $\frac{\partial}{\partial y} f(x, y)$ we will need to apply the chain rule once again, and so:

(4)
\begin{align} \quad \frac{\partial}{\partial y} f(x, y) = 6(1 + xy)^5 x \end{align}

## Example 3

Let $f(x, y) = \ln (x^2 + y^4 + 1)$. Find $\frac{\partial}{\partial x} f(x,y)$ and $\frac{\partial}{\partial y} f(x,y)$.

To calculate $\frac{\partial}{\partial x} f(x,y)$ we will need to apply the rule for differentiating logarithmic functions:

(5)
\begin{align} \quad \frac{\partial}{\partial x} f(x,y) = \frac{2x}{x^2 + y^4 + 1} \end{align}

To calculate $\frac{\partial}{\partial y} f(x,y)$ we will need to apply the rule for differentiating logarithmic functions once again:

(6)
\begin{align} \quad \frac{\partial}{\partial y} f(x,y) = \frac{4y^3}{x^2 + y^4 + 1} \end{align}

## Example 4

Let $f(x, y) = \frac{x^2 - y^2}{x^3 + y^3}$. Find $\frac{\partial}{\partial x} f(x,y)$ and $\frac{\partial}{\partial y} f(x,y)$.

To calculate $\frac{\partial}{\partial x} f(x,y)$ we will need to apply the quotient rule:

(7)
\begin{align} \frac{\partial}{\partial x} f(x, y) = \frac{(x^3 + y^3)(2x) - (x^2 - y^2)(3x^2)}{(x^3 + y^3)^2} \end{align}

To calculate $\frac{\partial}{\partial y} f(x,y)$ we will also need to apply the quotient rule:

(8)
\begin{align} \frac{\partial}{\partial y} f(x, y) = \frac{(x^3 + y^3)(-2y) - (x^2 - y^2)(3y^2)}{(x^3 + y^3)^2} \end{align}