Answer :
To solve the given problem, let's break it down into several steps.
1. Find [tex]\( f'(x) \)[/tex] - the derivative of [tex]\( f(x) \)[/tex] using the quotient rule.
2. Evaluate the derivative at [tex]\( a = \frac{\pi}{4} \)[/tex] to find the slope of the tangent line.
3. Find the value of the function at [tex]\( a = \frac{\pi}{4} \)[/tex].
4. Find the equation of the tangent line using the point-slope form of a line.
Let's start from the top.
### Step 1: Finding [tex]\( f'(x) \)[/tex]
Given the function:
[tex]\[ f(x) = \frac{6 \sin x}{2 \sin x + 4 \cos x} \][/tex]
We use the quotient rule for differentiation:
[tex]\[ f'(x) = \frac{u'v - uv'}{v^2} \][/tex]
where [tex]\( u = 6 \sin x \)[/tex] and [tex]\( v = 2 \sin x + 4 \cos x \)[/tex].
First, we find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = 6 \cos x \][/tex]
[tex]\[ v' = 2 \cos x - 4 \sin x \][/tex]
Applying the quotient rule:
[tex]\[ f'(x) = \frac{(6 \cos x)(2 \sin x + 4 \cos x) - (6 \sin x)(2 \cos x - 4 \sin x)}{(2 \sin x + 4 \cos x)^2} \][/tex]
Simplify the numerator:
[tex]\[ \text{Numerator} = 6 \cos x (2 \sin x + 4 \cos x) - 6 \sin x (2 \cos x - 4 \sin x) \][/tex]
[tex]\[ = 12 \cos x \sin x + 24 \cos^2 x - 12 \sin x \cos x + 24 \sin^2 x \][/tex]
[tex]\[ = 24 \cos^2 x + 24 \sin^2 x \][/tex]
[tex]\[ = 24 (\cos^2 x + \sin^2 x) \][/tex]
[tex]\[ = 24 \][/tex]
Note: The actual derived formula from the solution is a symbolic representation. Here we have:
[tex]\[ f'(x) = \frac{6 \cos x (2 \sin x + 4 \cos x) + 6 (4 \sin x - 2 \cos x) \sin x}{(2 \sin x + 4 \cos x)^2} \][/tex]
Thus:
[tex]\[ f'(x) = \frac{6\cos(x)}{2\sin(x) + 4\cos(x)} + \frac{6(4\sin(x) - 2\cos(x))\sin(x)}{(2\sin(x) + 4\cos(x))^2} \][/tex]
### Step 2: Evaluate the derivative at [tex]\( a = \frac{\pi}{4} \)[/tex]
We need to find [tex]\( f'(\pi/4) \)[/tex].
From the result:
[tex]\[ f'(\frac{\pi}{4}) = 1.33333333333333 \][/tex]
So, the slope [tex]\( m \)[/tex] of the tangent line at [tex]\( x = \frac{\pi}{4} \)[/tex]:
[tex]\[ m = 1.33333333333333 \][/tex]
### Step 3: Find [tex]\( f\left(\frac{\pi}{4}\right) \)[/tex]
Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = \frac{\pi}{4} \)[/tex]:
[tex]\[ f(\frac{\pi}{4}) = \frac{6 \sin(\frac{\pi}{4})}{2 \sin(\frac{\pi}{4}) + 4 \cos(\frac{\pi}{4})} \][/tex]
[tex]\[ = \frac{6 \cdot \frac{\sqrt{2}}{2}}{2 \cdot \frac{\sqrt{2}}{2} + 4 \cdot \frac{\sqrt{2}}{2}} \][/tex]
[tex]\[ = \frac{6 \cdot \frac{\sqrt{2}}{2}}{ \frac{2\sqrt{2}}{2} + \frac{4\sqrt{2}}{2}} \][/tex]
[tex]\[ = \frac{6 \cdot \frac{\sqrt{2}}{2}}{ \sqrt{2} + 2\sqrt{2}} \][/tex]
[tex]\[ = \frac{3\sqrt{2}}{ 3\sqrt{2}} = 1 \][/tex]
So, the value [tex]\( f \left( \frac{\pi}{4} \right) = 1 \)[/tex].
### Step 4: Find the equation of the tangent line
We use the point-slope form of a line equation:
[tex]\[ y - y_1 = m (x - x_1) \][/tex]
Where [tex]\( m = 1.33333333333333 \)[/tex], [tex]\( x_1 = \frac{\pi}{4} \)[/tex], and [tex]\( y_1 = 1 \)[/tex].
Rearrange to the slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y = 1.33333333333333 x + b \][/tex]
Now, calculate [tex]\( b \)[/tex]:
[tex]\[ 1 = 1.33333333333333 \cdot \frac{\pi}{4} + b \][/tex]
[tex]\[ b = 1 - 1.33333333333333 \cdot \frac{\pi}{4} \][/tex]
From the result, [tex]\( b = 1.0 - 0.333333333333333\pi \)[/tex].
Putting it together:
- The slope [tex]\( m = 1.33333333333333 \)[/tex].
- The y-intercept [tex]\( b = 1.0 - 0.333333333333333\pi \)[/tex].
Thus, the equation of the tangent line is:
[tex]\[ y = 1.33333333333333x + 1.0 - 0.333333333333333\pi \][/tex]
1. Find [tex]\( f'(x) \)[/tex] - the derivative of [tex]\( f(x) \)[/tex] using the quotient rule.
2. Evaluate the derivative at [tex]\( a = \frac{\pi}{4} \)[/tex] to find the slope of the tangent line.
3. Find the value of the function at [tex]\( a = \frac{\pi}{4} \)[/tex].
4. Find the equation of the tangent line using the point-slope form of a line.
Let's start from the top.
### Step 1: Finding [tex]\( f'(x) \)[/tex]
Given the function:
[tex]\[ f(x) = \frac{6 \sin x}{2 \sin x + 4 \cos x} \][/tex]
We use the quotient rule for differentiation:
[tex]\[ f'(x) = \frac{u'v - uv'}{v^2} \][/tex]
where [tex]\( u = 6 \sin x \)[/tex] and [tex]\( v = 2 \sin x + 4 \cos x \)[/tex].
First, we find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = 6 \cos x \][/tex]
[tex]\[ v' = 2 \cos x - 4 \sin x \][/tex]
Applying the quotient rule:
[tex]\[ f'(x) = \frac{(6 \cos x)(2 \sin x + 4 \cos x) - (6 \sin x)(2 \cos x - 4 \sin x)}{(2 \sin x + 4 \cos x)^2} \][/tex]
Simplify the numerator:
[tex]\[ \text{Numerator} = 6 \cos x (2 \sin x + 4 \cos x) - 6 \sin x (2 \cos x - 4 \sin x) \][/tex]
[tex]\[ = 12 \cos x \sin x + 24 \cos^2 x - 12 \sin x \cos x + 24 \sin^2 x \][/tex]
[tex]\[ = 24 \cos^2 x + 24 \sin^2 x \][/tex]
[tex]\[ = 24 (\cos^2 x + \sin^2 x) \][/tex]
[tex]\[ = 24 \][/tex]
Note: The actual derived formula from the solution is a symbolic representation. Here we have:
[tex]\[ f'(x) = \frac{6 \cos x (2 \sin x + 4 \cos x) + 6 (4 \sin x - 2 \cos x) \sin x}{(2 \sin x + 4 \cos x)^2} \][/tex]
Thus:
[tex]\[ f'(x) = \frac{6\cos(x)}{2\sin(x) + 4\cos(x)} + \frac{6(4\sin(x) - 2\cos(x))\sin(x)}{(2\sin(x) + 4\cos(x))^2} \][/tex]
### Step 2: Evaluate the derivative at [tex]\( a = \frac{\pi}{4} \)[/tex]
We need to find [tex]\( f'(\pi/4) \)[/tex].
From the result:
[tex]\[ f'(\frac{\pi}{4}) = 1.33333333333333 \][/tex]
So, the slope [tex]\( m \)[/tex] of the tangent line at [tex]\( x = \frac{\pi}{4} \)[/tex]:
[tex]\[ m = 1.33333333333333 \][/tex]
### Step 3: Find [tex]\( f\left(\frac{\pi}{4}\right) \)[/tex]
Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = \frac{\pi}{4} \)[/tex]:
[tex]\[ f(\frac{\pi}{4}) = \frac{6 \sin(\frac{\pi}{4})}{2 \sin(\frac{\pi}{4}) + 4 \cos(\frac{\pi}{4})} \][/tex]
[tex]\[ = \frac{6 \cdot \frac{\sqrt{2}}{2}}{2 \cdot \frac{\sqrt{2}}{2} + 4 \cdot \frac{\sqrt{2}}{2}} \][/tex]
[tex]\[ = \frac{6 \cdot \frac{\sqrt{2}}{2}}{ \frac{2\sqrt{2}}{2} + \frac{4\sqrt{2}}{2}} \][/tex]
[tex]\[ = \frac{6 \cdot \frac{\sqrt{2}}{2}}{ \sqrt{2} + 2\sqrt{2}} \][/tex]
[tex]\[ = \frac{3\sqrt{2}}{ 3\sqrt{2}} = 1 \][/tex]
So, the value [tex]\( f \left( \frac{\pi}{4} \right) = 1 \)[/tex].
### Step 4: Find the equation of the tangent line
We use the point-slope form of a line equation:
[tex]\[ y - y_1 = m (x - x_1) \][/tex]
Where [tex]\( m = 1.33333333333333 \)[/tex], [tex]\( x_1 = \frac{\pi}{4} \)[/tex], and [tex]\( y_1 = 1 \)[/tex].
Rearrange to the slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y = 1.33333333333333 x + b \][/tex]
Now, calculate [tex]\( b \)[/tex]:
[tex]\[ 1 = 1.33333333333333 \cdot \frac{\pi}{4} + b \][/tex]
[tex]\[ b = 1 - 1.33333333333333 \cdot \frac{\pi}{4} \][/tex]
From the result, [tex]\( b = 1.0 - 0.333333333333333\pi \)[/tex].
Putting it together:
- The slope [tex]\( m = 1.33333333333333 \)[/tex].
- The y-intercept [tex]\( b = 1.0 - 0.333333333333333\pi \)[/tex].
Thus, the equation of the tangent line is:
[tex]\[ y = 1.33333333333333x + 1.0 - 0.333333333333333\pi \][/tex]
