Question

Find both first partial derivatives. \(z=\cos \left(x^{2}+y^{2}\right)\)

Short answer

In this case, the first partial derivatives of the function are \(z_x = -2x \sin \left(x^{2} + y^{2}\right)\) and \(z_y = -2y \sin \left(x^{2} + y^{2}\right)\)

Step-by-step solution

Compute the First Partial Derivative with respect to x

Hold \(y\) constant and differentiate the function with respect to \(x\). The chain rule applies due to the composition of the function. So the derivative of \( \cos(u) \) is \( -\sin(u) \), times the derivative of \(u\) with respect to \(x\). This gives: \[ z_x = \frac{ \partial }{ \partial x} \cos \left(x^{2} + y^{2}\right) = -2x \sin \left(x^{2} + y^{2}\right) \]

Compute the First Partial Derivative with respect to y

Hold \(x\) constant and differentiate the function with respect to \(y\). As before, the chain rule applies. So the derivative of \( \cos(u) \) is \( -\sin(u) \), times the derivative of \(u\) with respect to \(y\). This gives: \[ z_y = \frac{ \partial }{ \partial y} \cos \left(x^{2} + y^{2}\right) = -2y \sin \left(x^{2} + y^{2}\right) \]