Problem 10 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(x, y)=y^{8}+2 x^{6}+2 x y$$

Step-by-Step Solution

Answer

Question: Find the first partial derivatives of the function $$f(x, y) = y^{8} + 2x^{6} + 2xy$$ with respect to x and y. Answer: The first partial derivatives of the given function are $$\frac{\partial f(x, y)}{\partial x} = 12x^{5} + 2y$$ and $$\frac{\partial f(x, y)}{\partial y} = 8y^{7} + 2x$$.

Step 1: Identify the given function

The given function is: $$f(x, y)=y^{8}+2 x^{6}+2 x y$$

Step 2: Find the partial derivative with respect to x

To find the partial derivative with respect to x, treat y as a constant and differentiate the function f(x, y) with respect to x: $$\frac{\partial f(x, y)}{\partial x} = \frac{\partial}{\partial x}\left(y^{8}+2 x^{6}+2 x y\right)$$

Step 3: Apply the power rule and constant rule on the first partial derivative

Use the power rule and constant rule to differentiate each term: $$\frac{\partial f(x, y)}{\partial x} = 0+12x^{5}+2y$$ So the first partial derivative with respect to x is: $$\frac{\partial f(x, y)}{\partial x} = 12x^{5} + 2y$$

Step 4: Find the partial derivative with respect to y

To find the partial derivative with respect to y, treat x as a constant and differentiate the function f(x, y) with respect to y: $$\frac{\partial f(x, y)}{\partial y} = \frac{\partial}{\partial y}\left(y^{8}+2 x^{6}+2 x y\right)$$

Step 5: Apply the power rule and constant rule on the second partial derivative

Use the power rule and constant rule to differentiate each term: $$\frac{\partial f(x, y)}{\partial y} = 8y^{7}+0+2x$$ So the first partial derivative with respect to y is: $$\frac{\partial f(x, y)}{\partial y} = 8y^{7} + 2x$$ The first partial derivatives of the given function are: $$\frac{\partial f(x, y)}{\partial x} = 12x^{5} + 2y \ and \ \frac{\partial f(x, y)}{\partial y} = 8y^{7} + 2x$$