find all second order derivatives for ( z = 7ye^{6x} ).\n( z_{xx}=252ye^{6x} )\n( z_{yy}=0 )\n(…

Explanation:

Step1: Find the first - order partial derivative (z_x)

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) (where (u = 7y) and (v = e^{6x})). Since (\frac{\partial(7y)}{\partial x}=0) and (\frac{\partial(e^{6x})}{\partial x}=6e^{6x}), we have (z_x=\frac{\partial z}{\partial x}=7y\times6e^{6x}=42ye^{6x})

Step2: Find the first - order partial derivative (z_y)

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) (where (u = 7y) and (v = e^{6x})). Since (\frac{\partial(7y)}{\partial y}=7) and (\frac{\partial(e^{6x})}{\partial y}=0), we have (z_y=\frac{\partial z}{\partial y}=7e^{6x})

Step3: Find (z_{xy})

Differentiate (z_x = 42ye^{6x}) with respect to (y). Using the rule (\frac{\partial(uv)}{\partial y}=u^\prime v+uv^\prime) (where (u = 42y) and (v = e^{6x})), (\frac{\partial(42y)}{\partial y}=42) and (\frac{\partial(e^{6x})}{\partial y}=0). So (z_{xy}=42e^{6x})

Step4: Find (z_{yx})

Differentiate (z_y = 7e^{6x}) with respect to (x). Using the rule (\frac{\partial(ae^{bx})}{\partial x}=abe^{bx}) (where (a = 7) and (b = 6)), we have (z_{yx}=7\times6e^{6x}=42e^{6x})

Answer:

(z_{xy}=z_{yx}=42e^{6x})