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Let's find the equation of the tangent line to the graph of the function [tex]\( f(x) = 4x \sin(x) \)[/tex] at the point [tex]\(\left(\frac{\pi}{2}, 2\pi\right)\)[/tex].
1. Find the derivative of the function [tex]\( f(x) \)[/tex]:
The function is [tex]\( f(x) = 4x \sin(x) \)[/tex].
To find the derivative [tex]\( f'(x) \)[/tex], we use the product rule, which states that if [tex]\( f(x) = u(x)v(x) \)[/tex], then [tex]\( f'(x) = u'(x)v(x) + u(x)v'(x) \)[/tex].
Here [tex]\( u(x) = 4x \)[/tex] and [tex]\( v(x) = \sin(x) \)[/tex].
- [tex]\( u'(x) = 4 \)[/tex]
- [tex]\( v'(x) = \cos(x) \)[/tex]
Applying the product rule:
[tex]\[ f'(x) = u'(x)v(x) + u(x)v'(x) = 4 \sin(x) + 4x \cos(x) \][/tex]
So, the derivative of the function is:
[tex]\[ f'(x) = 4 \sin(x) + 4x \cos(x) \][/tex]
2. Evaluate the derivative at the specified point [tex]\( x = \frac{\pi}{2} \)[/tex]:
We need to find [tex]\( f'\left(\frac{\pi}{2}\right) \)[/tex].
[tex]\[ f'\left(\frac{\pi}{2}\right) = 4 \sin\left(\frac{\pi}{2}\right) + 4 \left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) \][/tex]
Knowing that [tex]\( \sin\left(\frac{\pi}{2}\right) = 1 \)[/tex] and [tex]\( \cos\left(\frac{\pi}{2}\right) = 0 \)[/tex]:
[tex]\[ f'\left(\frac{\pi}{2}\right) = 4 \cdot 1 + 4 \cdot \frac{\pi}{2} \cdot 0 = 4 \][/tex]
Therefore, the slope [tex]\( m \)[/tex] of the tangent line at [tex]\( x = \frac{\pi}{2} \)[/tex] is [tex]\( 4 \)[/tex].
3. Use the point-slope form of a line to find the equation of the tangent line:
The tangent line passes through the point [tex]\(\left(\frac{\pi}{2}, 2\pi\right)\)[/tex], and has a slope of [tex]\( 4 \)[/tex].
The point-slope form of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the point on the line.
Substituting [tex]\( m = 4 \)[/tex], [tex]\( x_1 = \frac{\pi}{2} \)[/tex], and [tex]\( y_1 = 2\pi \)[/tex]:
[tex]\[ y - 2\pi = 4 \left( x - \frac{\pi}{2} \right) \][/tex]
4. Simplify the equation to the slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y - 2\pi = 4x - 2\pi \][/tex]
Adding [tex]\( 2\pi \)[/tex] to both sides:
[tex]\[ y = 4x \][/tex]
Thus, the equation of the tangent line to the graph of the function [tex]\( f(x) = 4x \sin(x) \)[/tex] at the point [tex]\(\left(\frac{\pi}{2}, 2\pi\right)\)[/tex] is:
[tex]\[ y = 4x \][/tex]
1. Find the derivative of the function [tex]\( f(x) \)[/tex]:
The function is [tex]\( f(x) = 4x \sin(x) \)[/tex].
To find the derivative [tex]\( f'(x) \)[/tex], we use the product rule, which states that if [tex]\( f(x) = u(x)v(x) \)[/tex], then [tex]\( f'(x) = u'(x)v(x) + u(x)v'(x) \)[/tex].
Here [tex]\( u(x) = 4x \)[/tex] and [tex]\( v(x) = \sin(x) \)[/tex].
- [tex]\( u'(x) = 4 \)[/tex]
- [tex]\( v'(x) = \cos(x) \)[/tex]
Applying the product rule:
[tex]\[ f'(x) = u'(x)v(x) + u(x)v'(x) = 4 \sin(x) + 4x \cos(x) \][/tex]
So, the derivative of the function is:
[tex]\[ f'(x) = 4 \sin(x) + 4x \cos(x) \][/tex]
2. Evaluate the derivative at the specified point [tex]\( x = \frac{\pi}{2} \)[/tex]:
We need to find [tex]\( f'\left(\frac{\pi}{2}\right) \)[/tex].
[tex]\[ f'\left(\frac{\pi}{2}\right) = 4 \sin\left(\frac{\pi}{2}\right) + 4 \left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) \][/tex]
Knowing that [tex]\( \sin\left(\frac{\pi}{2}\right) = 1 \)[/tex] and [tex]\( \cos\left(\frac{\pi}{2}\right) = 0 \)[/tex]:
[tex]\[ f'\left(\frac{\pi}{2}\right) = 4 \cdot 1 + 4 \cdot \frac{\pi}{2} \cdot 0 = 4 \][/tex]
Therefore, the slope [tex]\( m \)[/tex] of the tangent line at [tex]\( x = \frac{\pi}{2} \)[/tex] is [tex]\( 4 \)[/tex].
3. Use the point-slope form of a line to find the equation of the tangent line:
The tangent line passes through the point [tex]\(\left(\frac{\pi}{2}, 2\pi\right)\)[/tex], and has a slope of [tex]\( 4 \)[/tex].
The point-slope form of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the point on the line.
Substituting [tex]\( m = 4 \)[/tex], [tex]\( x_1 = \frac{\pi}{2} \)[/tex], and [tex]\( y_1 = 2\pi \)[/tex]:
[tex]\[ y - 2\pi = 4 \left( x - \frac{\pi}{2} \right) \][/tex]
4. Simplify the equation to the slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y - 2\pi = 4x - 2\pi \][/tex]
Adding [tex]\( 2\pi \)[/tex] to both sides:
[tex]\[ y = 4x \][/tex]
Thus, the equation of the tangent line to the graph of the function [tex]\( f(x) = 4x \sin(x) \)[/tex] at the point [tex]\(\left(\frac{\pi}{2}, 2\pi\right)\)[/tex] is:
[tex]\[ y = 4x \][/tex]
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