Partial derivative examples

Example 1

Let f (x, y) = y³x². Calculate $${\frac{{\partial f}}{{\partial x}}}$$(x, y).

Solution: To calculate $${\frac{{\partial f}}{{\partial x}}}$$(x, y), we simply view y as being a fixed number and calculate the ordinary derivative with respect to x. The first time you do this, it might be easiest to set y = b, where b is a constant, to remind you that you should treat y as though it were number rather than a variable. Then, the partial derivative $${\frac{{\partial f}}{{\partial x}}}$$(x, y) is the same as the ordinary derivative of the function g(x) = b³x². Using the rules for ordinary differentiation, we know that

$${\frac{{\mathrm{d} g}}{{\mathrm{d} x}}}$$

Now, we remember that b = y and substitute y back in to conclude that

$${\frac{{\partial f}}{{\partial x}}}$$

Example 2

For the same f, calculate $${\frac{{\partial f}}{{\partial y}}}$$(x, y).

Solution: This time, we'll just calculate the derivative with respect to y directly without replacing x with a constant. We just have to remember to treat x like a constant and use the rules for ordinary differentiation. We don't touch the x² and only differentiate the y³ factor to calculate that

$${\frac{{\partial f}}{{\partial y}}}$$

Example 3

For the same f, calculate $${\frac{{\partial f}}{{\partial x}}}$$(1, 2).

Solution: From example 1, we know that $${\frac{{\partial f}}{{\partial x}}}$$(x, y) = 2y³x. To evaluate this partial derivative at the point (x, y) = (1, 2), we just substitute the respective values for x and y:

$${\frac{{\partial f}}{{\partial x}}}$$

Example 4

For

$${\frac{{\cos(x_1x_4)\sin(x_2^5)}}{{e^{x_2}+ (1+x_2^2)/(x_1x_2x_4)}}}$$

calculate $${\frac{{\partial f}}{{\partial x_3}}}$$(a, b, c, d ).

Solution: Although this initially looks hard, it's really any easy problem. The ugly term does not depend on x₃, so in calculating partial derivative with respect to x₃, we treat it as a constant. The derivative of a constant is zero, so that term drops out. The derivative is just the derivative of the last term with respect to x₃, which is

$${\frac{{\partial f}}{{\partial x_3}}}$$

Substituting in the values (x₁, x₂, x₃, x₄) = (a, b, c, d ), we obtain the final answer

$${\frac{{\partial f}}{{\partial x_3}}}$$