Question

find all second order derivatives for ( z = 7ye^{6x} ).
( z_{xx}=252ye^{6x} )
( z_{yy}=0 )
( z_{xy}=z_{yx}=square )

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:

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