Partial Derivatives of 5x^2 y - 3x + 4y - 7

Sort the terms before you differentiate

Four terms, each with a different mix of variables. Sorting them first makes both partials nearly automatic.

• $5x^2y$ contains both variables, so it survives in both partials.
• $-3x$ contains only $x$, so it survives only in $\frac{\partial f}{\partial x}$, as $-3$.
• $4y$ contains only $y$, so it survives only in $\frac{\partial f}{\partial y}$, as $4$.
• $-7$ contains neither, so it vanishes from both.

At $(2,-1)$ these give $10(2)(-1) - 3 = -23$ and $5(4) + 4 = 24$.

The mistake: keeping the constant, or the wrong linear term

Answers like $10xy - 3 - 7$ show a constant that should have been differentiated away. The derivative of any constant is zero, in one variable or several.

The subtler error is treating $4y$ as a constant in $\frac{\partial f}{\partial y}$ because it looks like a leftover, or carrying $-3x$ into the $y$ partial. During $\frac{\partial f}{\partial y}$, the term $-3x$ genuinely is a constant and contributes $0$, while $4y$ is genuinely variable and contributes $4$.

One line of checking catches all of this. Set $x$ to a specific number, say $x = 2$, and read off the resulting one-variable function: $f(2,y) = 20y - 6 + 4y - 7 = 24y - 13$. Its derivative is $24$, which agrees with $5(2)^2 + 4$.

A gradient that never vanishes

The gradient is $\nabla f = \left\langle 10xy - 3,\; 5x^2 + 4 \right\rangle$. Its second component is at least $4$ for every real $x$, so it can never be zero.

That means this surface has no critical points at all: no maxima, no minima, no saddles. Wherever you stand, there is a direction of strict increase.

The gradient is also what you need for linearisation. At $(2,-1)$, where $f = -20 - 6 - 4 - 7 = -37$, the tangent plane is the following.