Problem 61 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Calculate the first partial derivatives of the function \(f(x, y) = 1 - \tan^{-1}(x^2 + y^2)\). Answer: The first partial derivatives of the function are: $$f_x = -\frac{2x}{1 + (x^2 + y^2)^2}$$ $$f_y = -\frac{2y}{1 + (x^2 + y^2)^2}$$

Step 1: Recall the chain rule

For a function \(f(x, y)\), the chain rule states that: $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial g} \cdot \frac{\partial g}{\partial x}$$ where \(g\) is an inner function of \(f\). We will apply the chain rule to compute both partial derivatives.

Step 2: Compute \(f_x\)

Let's first compute the partial derivative of f with respect to x. We have \(g(x, y) = x^2 + y^2\), so \(f(x, y) = 1 - \tan^{-1}(g(x, y))\). Now applying the chain rule: $$f_x = \frac{\partial f}{\partial g} \cdot \frac{\partial g}{\partial x}$$ Find the derivative of \(f\) with respect to \(g\): $$\frac{\partial f}{\partial g} = - \frac{1}{1 + g^2}$$ Find the derivative of \(g\) with respect to \(x\): $$\frac{\partial g}{\partial x} = 2x$$ Substitute and compute the product: $$f_x = - \frac{1}{1 + (x^2 + y^2)^2} \cdot 2x$$ So the first partial derivative with respect to x is: $$f_x = -\frac{2x}{1 + (x^2 + y^2)^2}$$

Step 3: Compute \(f_y\)

Now let's compute the partial derivative of f with respect to y. Again, we have \(g(x, y) = x^2 + y^2\), so \(f(x, y) = 1 - \tan^{-1}(g(x, y))\). Applying the chain rule: $$f_y = \frac{\partial f}{\partial g} \cdot \frac{\partial g}{\partial y}$$ We already found the derivative of \(f\) with respect to \(g\) in Step 2: $$\frac{\partial f}{\partial g} = - \frac{1}{1 + g^2}$$ Now find the derivative of \(g\) with respect to \(y\): $$\frac{\partial g}{\partial y} = 2y$$ Substitute and compute the product: $$f_y = - \frac{1}{1 + (x^2 + y^2)^2} \cdot 2y$$ So the first partial derivative with respect to y is: $$f_y = -\frac{2y}{1 + (x^2 + y^2)^2}$$

Step 4: Write down the final answer

The first partial derivatives of the function \(f(x, y) = 1 - \tan^{-1}(x^2 + y^2)\) are: $$f_x = -\frac{2x}{1 + (x^2 + y^2)^2}$$ $$f_y = -\frac{2y}{1 + (x^2 + y^2)^2}$$