Problem 7 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(x, y)=3 x^{2}+4 y^{3}$$

Step-by-Step Solution

Answer

Answer: The first partial derivatives of the function are $$\frac{\partial f}{\partial x}=6x$$ and $$\frac{\partial f}{\partial y}=12y^2$$.

Step 1: Find the partial derivative with respect to x

To find the partial derivative with respect to x, we take the derivative of the function treating y as a constant. It can be represented as $$\frac{\partial f}{\partial x}$$. We use the power rule for differentiation: $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial}{\partial x}(3x^2+4y^3)$$ Remember that anything without the x variable is treated as a constant during this process. $$\frac{\partial f(x,y)}{\partial x}=(2\cdot3)x^{(2-1)}+0$$ $$\frac{\partial f(x,y)}{\partial x}=6x$$ So, the first partial derivative of the function with respect to x is $$\frac{\partial f}{\partial x}=6x$$.

Step 2: Find the partial derivative with respect to y

To find the partial derivative with respect to y, we take the derivative of the function treating x as a constant. It can be represented as $$\frac{\partial f}{\partial y}$$. We use the power rule for differentiation: $$\frac{\partial f(x,y)}{\partial y}=\frac{\partial}{\partial y}(3x^2+4y^3)$$ Remember that anything without the y variable is treated as a constant during this process. $$\frac{\partial f(x,y)}{\partial y}=0+(3\cdot4)y^{(3-1)}$$ $$\frac{\partial f(x,y)}{\partial y}=12y^2$$ So, the first partial derivative of the function with respect to y is $$\frac{\partial f}{\partial y}=12y^2$$.

Step 3: Final Answer

The first partial derivatives of the given function $$f(x, y)=3 x^{2}+4 y^{3}$$ are: $$\frac{\partial f}{\partial x}=6x$$ $$\frac{\partial f}{\partial y}=12y^2$$