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Given z = x2+y2 + x2+y3, Evaluate (i) (dz)/(dx) and (ii) (dz)/(dy) when x=1, and y=1.

August 20, 2026Mathematics

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Answers

2 parts

4x

6

3 steps

Here's the solution to the problem.

Given the function $z = x^2 + y^3 + x^2 + y^3$ .

Step 1: Simplify the expression for $z$ .

z = 2x^2 + 2y^3

Evaluate $\frac{dz}{dx}$ .

Differentiate $z$ with respect to $x$ , treating $y$ as a constant.
$\frac{dz}{dx} = \frac{\partial}{\partial x}(2x^2 + 2y^3)$ $\frac{dz}{dx} = \frac{\partial}{\partial x}(2x^2) + \frac{\partial}{\partial x}(2y^3)$ $\frac{dz}{dx} = 4x + 0$ $\frac{dz}{dx} = 4x$
The partial derivative $\frac{dz}{dx}$ is $\boxed{4x}$ .

Evaluate $\frac{dz}{dy}$ when $x=1$ and $y=1$ .

Differentiate $z$ with respect to $y$ , treating $x$ as a constant.
$\frac{dz}{dy} = \frac{\partial}{\partial y}(2x^2 + 2y^3)$ $\frac{dz}{dy} = \frac{\partial}{\partial y}(2x^2) + \frac{\partial}{\partial y}(2y^3)$ $\frac{dz}{dy} = 0 + 6y^2$ $\frac{dz}{dy} = 6y^2$
Substitute $x=1$ and $y=1$ into the expression for $\frac{dz}{dy}$ .
$\frac{dz}{dy} \Big|_{(x=1, y=1)} = 6(1)^2$ $\frac{dz}{dy} \Big|_{(x=1, y=1)} = 6(1)$ $\frac{dz}{dy} \Big|_{(x=1, y=1)} = 6$
The partial derivative $\frac{dz}{dy}$ at $x=1, y=1$ is $\boxed{6}$ .

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