Questions: Differentiate the following function.
y=(9+sin(x))/(9x+cos(x))
y'=
Transcript text: Differentiate the following function.
\[
\begin{array}{r}
y=\frac{9+\sin (x)}{9 x+\cos (x)} \\
y^{\prime}=\square
\end{array}
\]
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Solution
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 \( v(x) = 9x + \cos(x) \).
Step 1: Identify the Functions
Given the function \( y = \frac{9 + \sin(x)}{9x + \cos(x)} \), we identify:
\( u(x) = 9 + \sin(x) \)
\( v(x) = 9x + \cos(x) \)
Step 2: Differentiate \( u(x) \) and \( v(x) \)
Calculate the derivatives:
\( u'(x) = \cos(x) \)
\( v'(x) = 9 - \sin(x) \)
Step 3: Apply the Quotient Rule
Using the quotient rule:
\[
y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]
Substitute the derivatives and functions:
\[
y' = \frac{\cos(x)(9x + \cos(x)) - (9 + \sin(x))(9 - \sin(x))}{(9x + \cos(x))^2}
\]