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