(v) y = cos^-1(sin5x)

To solve the problem, I'll use the chain rule and the trigonometric identities.

Python

import sympy as sp

# Define the symbols
x = sp.symbols('x')
y = sp.acos(sp.sin(5*x))

# Calculate the derivative
dy_dx = sp.diff(y, x)

# Simplify the derivative
dy_dx = sp.simplify(dy_dx)

# Print the result
print("The derivative of y = cos^-1(sin5x) is:")
print(dy_dx)

Code output

The derivative of y = cos^-1(sin5x) is:
-5*cos(5*x)/sqrt(cos(5*x)**2)

The derivative of $y={\cos }^{-1}(\sin 5x)$ is:

$\frac{dy}{dx}=\frac{-5\cos (5x)}{\sqrt{{\cos }^2(5x)}}$