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Abdessamad
AbdessamadVerified by MethodMath staff
May 16, 2026

Partial vs total derivatives in multivariable calculus

In multivariable calculus, I see both partial derivatives ∂f∂x\frac{\partial f}{\partial x}∂x∂f​ and total derivatives dfdx\frac{df}{dx}dxdf​. When should I use each?

For a function f(x,y)f(x, y)f(x,y) where y=y(x)y = y(x)y=y(x), the chain rule gives:

dfdx=∂f∂x+∂f∂ydydx\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}dxdf​=∂x∂f​+∂y∂f​dxdy​

What does this mean intuitively? Can someone provide a concrete example where mixing them up leads to wrong results?

1 answers701 views

1 Answer

2
James Chen
May 16, 2026
Accepted
The key distinction: - **Partial derivative** $\frac{\partial f}{\partial x}$ holds all other independent variables constant. It tells you how $f$ changes when you wiggle $x$ while keeping $y$ fixed. - **Total derivative** $\frac{df}{dx}$ accounts for all paths through which changing $x$ affects $f$, including indirect paths via other variables that depend on $x$. **Concrete example:** Suppose $f(x, y) = x^2 + y^2$ and $y = 3x$. Then $\frac{\partial f}{\partial x} = 2x$ (treating $y$ as constant). But $\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx} = 2x + 2y \cdot 3 = 2x + 6y = 2x + 6(3x) = 20x$. We can verify: substituting $y = 3x$ gives $f(x) = x^2 + (3x)^2 = 10x^2$, and indeed $\frac{d}{dx}(10x^2) = 20x$. **When to use which:** Use partial derivatives for multivariate functions where variables are truly independent. Use total derivatives when variables are related (e.g., in physics along a path, or in the chain rule for composite functions).
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