Computing Partial Derivatives

The ordinary rules, one variable at a time

A partial derivative needs no new rules. To find $\frac{\partial f}{\partial x}$∂f/∂x, treat every other variable as a constant, a fixed number, and differentiate in x with the power rule, the product rule, the quotient rule and the chain rule exactly as before.

Take $f\left(x,y\right)=x^{2}y+y^3$f(x, y) = x²y + y³. In $x^{2}y$x²y, the y is a constant factor, so it stays, and $x^2$x² becomes 2x by the power rule: the term gives 2xy. The term $y^3$y³ has no x in it, so it is a constant and gives 0. So $\frac{\partial f}{\partial x}=2xy$∂f/∂x = 2xy, which is 4 at (1, 2).

Three slices of $z=x^{2}y$z = x²y, each with y held at a constant value: $z=x^2$z = x² at y = 1 and $z=3x^2$z = 3x² at y = 3, dashed, and $z=2x^2$z = 2x² at y = 2, in gold. At x = 1 their gradients are 2, 4 and 6, which is 2xy each time. The constant y multiplies the slice, and so it multiplies the slope.

Swap the roles

To find $\frac{\partial f}{\partial y}$∂f/∂y, hold x constant instead and differentiate in y. In $x^{2}y$x²y, the $x^2$x² is now the constant factor and y differentiates to 1, so the term gives $x^2$x². The term $y^3$y³ gives $3y^2$3y². So $\frac{\partial f}{\partial y}=x^2+3y^2$∂f/∂y = x² + 3y², which is 1 + 12 = 13 at (1, 2).

Check it with a small step in y: f(1, 2) = 10 and f(1, 2.001) = 10.013006, so the difference quotient is $\frac{0.013006}{0.001}=13.006$0.013006/0.001 = 13.006, close to 13. A step of 0.001 in x gives 4.002, close to the 4 found for $\frac{\partial f}{\partial x}$∂f/∂x.

A term without the variable

A term that does not contain the variable you differentiate with respect to is a constant, so its partial derivative is 0. This is the same as the derivative of a number being 0.

For $f\left(x,y\right)=3x^2+4xy-7y^2$f(x, y) = 3x² + 4xy − 7y², the term $-7y^2$−7y² vanishes under $\frac{\partial }{\partial x}$∂/∂x and $3x^2$3x² vanishes under $\frac{\partial }{\partial y}$∂/∂y. So $\frac{\partial f}{\partial x}=6x+4y$∂f/∂x = 6x + 4y and $\frac{\partial f}{\partial y}=4x-14y$∂f/∂y = 4x − 14y. At (1, 2) these are 6 + 8 = 14 and 4 − 28 = −24.

A term can be a constant to one partial derivative and not to the other. In 4xy, both variables appear, so it contributes 4y to $\frac{\partial f}{\partial x}$∂f/∂x and 4x to $\frac{\partial f}{\partial y}$∂f/∂y.

Products, quotients and the chain rule

When the variable appears in two factors, the product rule is needed. For $f\left(x,y\right)=x{e}^{xy}$f(x, y) = x e^(xy), x appears in both x and $e^{xy}$e^(xy). In x, the chain rule gives the derivative of $e^{xy}$e^(xy) as $y{e}^{xy}$y e^(xy), so $\frac{\partial f}{\partial x}=e^{xy}+xy{e}^{xy}=\left(1+xy\right)e^{xy}$∂f/∂x = e^(xy) + x y e^(xy) = (1 + xy)e^(xy). In y, x is a constant factor, so $\frac{\partial f}{\partial y}=x\cdot x{e}^{xy}=x^2{e}^{xy}$∂f/∂y = x · x e^(xy) = x² e^(xy). At (1, 0.5) these are $1.5e^{0.5}\approx 2.473$1.5e^(0.5) ≈ 2.473 and $e^{0.5}\approx 1.649$e^(0.5) ≈ 1.649.

For $f\left(x,y\right)=\sin \left(x^{2}y\right)$f(x, y) = sin(x²y), the chain rule brings out the derivative of the inside, $x^{2}y$x²y, in each variable: $\frac{\partial f}{\partial x}=2xy\cos \left(x^{2}y\right)$∂f/∂x = 2xy cos(x²y) and $\frac{\partial f}{\partial y}=x^2\cos \left(x^{2}y\right)$∂f/∂y = x² cos(x²y). At (1, 2) these are $4\cos 2\approx -1.665$4 cos 2 ≈ −1.665 and $\cos 2\approx -0.416$cos 2 ≈ −0.416.

For $f\left(x,y\right)=\frac{x}{x+y}$f(x, y) = x/(x + y), the quotient rule in x gives the numerator (x + y) × 1 − x × 1 = y over the denominator ${\left(x+y\right)}^2$(x + y)², so $\frac{\partial f}{\partial x}=\frac{y}{{\left(x+y\right)}^2}$∂f/∂x = y/(x + y)². In y the top x is a constant, so $\frac{\partial f}{\partial y}=-\frac{x}{{\left(x+y\right)}^2}$∂f/∂y = −x/(x + y)². At (1, 2) these are $\frac{2}{9}\approx 0.222$2/9 ≈ 0.222 and $-\frac{1}{9}\approx -0.111$−1/9 ≈ −0.111. Each of these values agrees with a difference quotient at the same point.

The usual mistakes

Deleting a constant factor. Under $\frac{\partial }{\partial x}$∂/∂x, the y in $3x^{2}y$3x²y is a constant factor and stays: the answer is 6xy, not 6x.

Differentiating a term that has no x in it. Under $\frac{\partial }{\partial x}$∂/∂x, $x^2+5y$x² + 5y gives 2x; the 5y is a constant and gives 0, not 5.

Mixing up the two partial derivatives. For $3x^{2}y$3x²y, $\frac{\partial f}{\partial x}=6xy$∂f/∂x = 6xy and $\frac{\partial f}{\partial y}=3x^2$∂f/∂y = 3x², and each answers a different question.

Leaving out the factor from the chain rule. The partial derivative of $e^{xy}$e^(xy) with respect to x is $y{e}^{xy}$y e^(xy), not $e^{xy}$e^(xy).