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Exercise 257Calculus › Multivariable calculus › Partial derivativesHigh schoolAdded 3 Nov 2026

Evaluating partial derivatives at a point

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Statement

Let f(x,y)=exy+ln⁡(1+x2)f(x,y)=e^{xy}+\ln(1+x^2). Compute ∂f∂x(0,1)\frac{\partial f}{\partial x}(0,1) and ∂f∂y(0,1)\frac{\partial f}{\partial y}(0,1).

Hint

Compute the partial derivatives, then substitute (x,y)=(0,1)(x,y)=(0,1).

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Solution

Partial derivatives.
∂f∂x=yexy+2x1+x2,∂f∂y=xexy.\frac{\partial f}{\partial x}=ye^{xy}+\frac{2x}{1+x^2},\qquad \frac{\partial f}{\partial y}=xe^{xy}.

At (0,1)(0,1). ∂f∂x(0,1)=1⋅1+0=1\frac{\partial f}{\partial x}(0,1)=1\cdot1+0=1, ∂f∂y(0,1)=0\frac{\partial f}{\partial y}(0,1)=0.

Conclusion. The values are 11 and 00.