Question

Find the first partial derivatives of the following functions. $$g(w, x, y, z)=\cos (w+x) \sin (y-z)$$

Short answer

Answer: The first partial derivatives of $$g(w, x, y, z)$$ are as follows: \[\frac{\partial g}{\partial w} = -\sin(w+x) \sin(y-z)\] \[\frac{\partial g}{\partial x} = -\sin(w+x) \sin(y-z)\] \[\frac{\partial g}{\partial y} = \cos(w+x) \cos(y-z)\] \[\frac{\partial g}{\partial z} = -\cos(w+x) \cos(y-z)\]

Step-by-step solution

Differentiate with respect to w

First, let's differentiate the function with respect to \(w\). During this step, we treat \(x, y\) and \(z\) as constants. We have: $$\frac{\partial g}{\partial w} = \frac{\partial}{\partial w} (\cos(w+x) \sin(y-z))$$ Since the function is the product of two functions, we apply the product rule: $$\frac{\partial g}{\partial w} = \frac{\partial (\cos(w+x))}{\partial w} \sin(y-z) + \cos(w+x) \frac{\partial(\sin(y-z))}{\partial w}$$ Now we apply the chain rule to the first term on the right side: $$\frac{\partial g}{\partial w} = -\sin(w+x) \sin(y-z) + \cos(w+x) (0)$$ Finally, simplify the expression: $$\frac{\partial g}{\partial w} = -\sin(w+x) \sin(y-z)$$

Differentiate with respect to x

Now we want to differentiate the function with respect to \(x\). Consider \(w, y\), and \(z\) to be constants. We proceed similarly as in Step 1: $$\frac{\partial g}{\partial x} = \frac{\partial}{\partial x} (\cos(w+x) \sin(y-z))$$ Using the product rule, and then applying the chain rule, we get: $$\frac{\partial g}{\partial x} = -\sin(w+x) \sin(y-z) + \cos(w+x) (0)$$ Simplifying the expression gives: $$\frac{\partial g}{\partial x} = -\sin(w+x) \sin(y-z)$$

Differentiate with respect to y

Now differentiate the function with respect to \(y\). Treat \(w, x\), and \(z\) as constants: $$\frac{\partial g}{\partial y} = \frac{\partial}{\partial y} (\cos(w+x) \sin(y-z))$$ Applying the product rule, and chain rule (when necessary), we obtain: $$\frac{\partial g}{\partial y} = \cos(w+x) \cos(y-z) + \cos(w+x) (0)$$ Simplifying the expression, we find: $$\frac{\partial g}{\partial y} = \cos(w+x) \cos(y-z)$$

Differentiate with respect to z

Finally, differentiate the function with respect to \(z\), treating \(w, x\), and \(y\) as constants: $$\frac{\partial g}{\partial z} = \frac{\partial}{\partial z} (\cos(w+x) \sin(y-z))$$ Utilizing the product rule and chain rule, we compute: $$\frac{\partial g}{\partial z} = \cos(w+x) (0) + \cos(w+x) (-\cos(y-z))$$ Simplifying the expression gives: $$\frac{\partial g}{\partial z} = -\cos(w+x) \cos(y-z)$$ So the four first partial derivatives of the function \(g(w,x,y,z)\) are: $$\frac{\partial g}{\partial w} = -\sin(w+x) \sin(y-z)$$ $$\frac{\partial g}{\partial x} = -\sin(w+x) \sin(y-z)$$ $$\frac{\partial g}{\partial y} = \cos(w+x) \cos(y-z)$$ $$\frac{\partial g}{\partial z} = -\cos(w+x) \cos(y-z)$$