Questions: Differentiate the following function. y=(9+sin(x))/(9x+cos(x)) y'=

Solution Steps

To differentiate the given function, we will use the quotient rule. The quotient rule states that if you have a function \( y = \frac{u(x)}{v(x)} \), then its derivative is given by \( y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Here, \( u(x) = 9 + \sin(x) \) and \( v(x) = 9x + \cos(x) \).

Given the function \( y = \frac{9 + \sin(x)}{9x + \cos(x)} \), we identify:

• \( u(x) = 9 + \sin(x) \)
• \( v(x) = 9x + \cos(x) \)

Calculate the derivatives:

• \( u'(x) = \cos(x) \)
• \( v'(x) = 9 - \sin(x) \)

Using the quotient rule: \[ y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

Substitute the derivatives and functions: \[ y' = \frac{\cos(x)(9x + \cos(x)) - (9 + \sin(x))(9 - \sin(x))}{(9x + \cos(x))^2} \]

Simplify the expression: \[ y' = \frac{9x\cos(x) + \cos^2(x) - (81 - 9\sin(x) + 9\sin(x) + \sin^2(x))}{(9x + \cos(x))^2} \]

Further simplification: \[ y' = \frac{9x\cos(x) + \cos^2(x) - 81 - \sin^2(x)}{(9x + \cos(x))^2} \]

Final Answer