please fill in the question marks:

# First we set the state of the network

$\backslash$sigma $=$ np$.$tanh

w$1=1\mathrm{.}3$

b$1=-0\mathrm{.}1$

# Then we define the neuron activation.

def a$1($a$0)$:

z $=$ w$1*$ a$0+$ b$1$

return $\backslash$sigma $($z$)$

# Experiment with different values of x below.

x $=0$

print$(($a$1($x$)-1)**2)$

# First define our sigma function.

sigma $=$ np$.$tanh

# Next define the feed$-$forward equation.

def a$1($w$1,$ b$1,$ a$0)$ :

z $=$ w$1*$ a$0+$ b$1$

return sigma$($z$)$

# The individual cost function is the square of the difference between

# the network output and the training data output.

def C $($w$1,$ b$1,$ x$,$ y$)$ :

return $($a$1($w$1,$ b$1,$ x$)-$ y$)**2$

# This function returns the derivative of the cost function with

# respect to the weight.

def dCdw $($w$1,$ b$1,$ x$,$ y$)$ :

z $=???$

dCda $=???$ # Derivative of cost with activation

dadz $=1/$np$.$cosh$($z$)**2$ # derivative of activation with weighted sum z

dzdw $=???$ # derivative of weighted sum z with weight

return $???$ # Return the chain rule product.

# This function returns the derivative of the cost function with

# respect to the bias.

# It is very similar to the previous function.

# You should complete this function.

def dCdb $($w$1,$ b$1,$ x$,$ y$)$ :

z $=???$

dCda $=???$

dadz $=???$

# Change the next line to give the derivative of

# the weighted sum, z$,$ with respect to the bias, b$.$

dzdb $=???$

return $???$

w$1=2\mathrm{.}3$

b$1=-1\mathrm{.}2$

# We can test on a single data point pair of x and y$.$

x $=0$

y $=1$

# Output how the cost would change

# in proportion to a small change in the bias

print$($ dCdb$($w$1,$ b$1,$ x$,$ y$))$

You should get $-1\mathrm{.}1186026425530913$