ESTION 1 Given \( z=\cos ^{2}\left(x^{2} y^{2}\right) \)

We are given \[ z=\cos^2\left(x^2y^2\right). \] Assuming the task is to compute the partial derivatives of \(z\) with respect to \(x\) and \(y\), we proceed as follows. ### Step 1. Rewrite the Function Write the function as \[ z=\left(\cos\left(x^2y^2\right)\right)^2. \] This shows that we have a composite function where the outer function is \[ f(u)=u^2 \] or, more directly, \[ f(v)=\cos^2(v) \] with the inner function \[ u=x^2y^2. \] ### Step 2. Differentiate the Outer Function Using the Chain Rule If we set \[ z=\cos^2(u), \] then differentiating with respect to \(u\) gives \[ \frac{dz}{du} = 2\cos(u)\cdot \left(-\sin(u)\right) = -2\cos(u)\sin(u). \] Using the double-angle identity \[ \sin(2u)=2\sin u\cos u, \] we can express the derivative as \[ \frac{dz}{du} = -\sin(2u). \] ### Step 3. Compute the Derivative of the Inner Function The inner function is \[ u=x^2y^2. \] Differentiate \(u\) with respect to \(x\) and \(y\) separately. #### Partial Derivative with Respect to \(x\) Treat \(y\) as a constant: \[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\left(x^2y^2\right) = 2xy^2. \] #### Partial Derivative with Respect to \(y\) Treat \(x\) as a constant: \[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}\left(x^2y^2\right) = 2x^2y. \] ### Step 4. Apply the Chain Rule to Find the Partial Derivatives of \(z\) #### Partial Derivative with Respect to \(x\) By the chain rule: \[ \frac{\partial z}{\partial x}= \frac{dz}{du}\cdot\frac{\partial u}{\partial x}. \] Substitute the results: \[ \frac{\partial z}{\partial x} = \left[-\sin\left(2u\right)\right]\cdot \left(2xy^2\right). \] Recall \(u=x^2y^2\); hence, \[ \frac{\partial z}{\partial x} = -2xy^2\sin\left(2x^2y^2\right). \] #### Partial Derivative with Respect to \(y\) Similarly, \[ \frac{\partial z}{\partial y}= \frac{dz}{du}\cdot \frac{\partial u}{\partial y}. \] Substitute: \[ \frac{\partial z}{\partial y} = \left[-\sin\left(2u\right)\right]\cdot \left(2x^2y\right), \] so that \[ \frac{\partial z}{\partial y} = -2x^2y\sin\left(2x^2y^2\right). \] ### Final Answers \[ \frac{\partial z}{\partial x} = -2xy^2\sin\left(2x^2y^2\right) \] \[ \frac{\partial z}{\partial y} = -2x^2y\sin\left(2x^2y^2\right) \]