Partial Derivatives of sin(xy)cos(z)

Split the function before you differentiate

This function factors into a part that depends only on $x$ and $y$ and a part that depends only on $z$. Whenever a function splits like that, each partial derivative differentiates one factor and copies the other down unchanged. No product rule is needed, because one of the two factors is constant for the variable in question.

For $\frac{\partial f}{\partial x}$, the factor $\cos(z)$ is a constant. Differentiate $\sin(xy)$ with respect to $x$: the outside derivative is $\cos(xy)$ and the inside derivative of $xy$ with respect to $x$ is $y$.

The $y$ partial is identical with the roles of $x$ and $y$ swapped, giving $x\cos(xy)\cos(z)$. For $\frac{\partial f}{\partial z}$ the factor $\sin(xy)$ is the constant, and the derivative of $\cos(z)$ is $-\sin(z)$.

The mistake: losing the inside derivative or the minus sign

The single most common error is writing $\cos(xy)\cos(z)$ for the $x$ partial and forgetting the factor $y$. The argument of the sine is $xy$, not $x$, so the chain rule owes you the inside derivative. A quick way to notice is dimensional: differentiating with respect to $x$ should leave $y$ present, and it does.

The second error is a sign. Only the $z$ partial picks up a minus, because only there do you differentiate a cosine. The $x$ and $y$ partials differentiate a sine, which gives a plain cosine with no sign change.

• Differentiating $\sin$ gives $+\cos$, with the inside derivative attached.
• Differentiating $\cos$ gives $-\sin$, with the inside derivative attached.
• The inside derivative of $xy$ is $y$ in the $x$ direction and $x$ in the $y$ direction.

A quick numerical spot check

Pick a point where the trigonometry is easy and test all three at once. At $(1,0,0)$ the argument $xy$ is $0$ and $z$ is $0$, so $\cos(xy) = 1$, $\sin(xy) = 0$, $\cos z = 1$ and $\sin z = 0$.

That gives $f_x = 0$, $f_y = 1$ and $f_z = 0$. The middle value makes sense: near that point, holding $x = 1$ and $z = 0$, the function behaves like $\sin(y)$, whose slope at $y = 0$ is exactly $1$. The other two vanish because moving in $x$ or $z$ from there does not change the value to first order.