Problem 62 - Calculus Early Transcendentals

Question

Compute the first partial derivatives of the following functions. $$f(x, y)=1-\cos (2(x+y))+\cos ^{2}(x+y)$$

Step-by-Step Solution

Answer

Question: Find the first partial derivatives of the function $$f(x, y) = 1 - \cos(2(x+y)) + \cos^2(x+y)$$ with respect to both x and y. Answer: The first partial derivatives of the function are given by: $$ \frac{\partial f}{\partial x} = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y) \\ \frac{\partial f}{\partial y} = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y) $$

Step 1: Compute the partial derivative with respect to x

Differentiate $$f(x,y)$$ with respect to $$x$$, considering $$y$$ as a constant. $$ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left(1 - \cos(2(x+y)) + \cos^2(x+y)\right) $$

Step 2: Apply chain rule for the cosine term

Apply chain rule for the $$\cos(2(x+y))$$ term: $$ \frac{\partial}{\partial x} \left[-\cos(2(x+y))\right] = -\sin(2(x+y)) \cdot \frac{\partial}{\partial x} (2(x+y)) $$ Now, differentiate $$2(x+y)$$ with respect to $$x$$: $$ \frac{\partial}{\partial x}(2(x+y)) = 2 $$ So, the derivative of the cosine term is: $$ -\sin(2(x+y)) \cdot 2 = -2\sin(2(x+y)) $$

Step 3: Apply chain rule for the cosine square term

Apply chain rule for the $$\cos^2(x+y)$$ term: $$ \frac{\partial}{\partial x} \left[\cos^2(x+y)\right] = 2\cos(x+y) \cdot \frac{\partial}{\partial x}(\cos(x+y)) $$ Now, differentiate $$\cos(x+y)$$ with respect to $$x$$: $$ \frac{\partial}{\partial x}(\cos(x+y)) = -\sin(x+y) $$ So, the derivative of the cosine square term is: $$ 2\cos(x+y) \cdot (-\sin(x+y)) = -2\cos(x+y)\sin(x+y) $$

Step 4: Combine the derivatives

The partial derivative with respect to $$x$$ is the sum of the derivatives from Steps 2 and 3, plus the derivative of the constant term (which is 0), so we have: $$ \frac{\partial f}{\partial x} = 0 - 2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y) $$

Step 5: Compute the partial derivative with respect to y

Differentiate $$f(x,y)$$ with respect to $$y$$, considering $$x$$ constant, in a similar way as we did for $$x$$: $$ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(1 - \cos(2(x+y)) + \cos^2(x+y)\right) = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y) $$ The first partial derivatives of the function $$f(x,y)$$ are: $$ \frac{\partial f}{\partial x} = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y) \\ \frac{\partial f}{\partial y} = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y) $$