Basically what I learned in the activity we did in class regarding tangent lines and derivatives was that the specific root of a function can be re-written as that function to the power of one over the specific root:
f(x)=√(x) -------> f(x)=x^(1/2)
Therefore, when derived for the function of the tangent line using the power rule, the equation becomes:
f'(x) = (1/2)x + b
After finding the y-intercept to be 1/2, you now have the equation of the tangent line:
f(x) = (1/2)x + 1/2
This is true for any other specific root of a function as well. If the function is transformed by being reflected over the x-axis (f(x)=-√(x)), the only thing that changes in the equation for the tangent line is the sign of the slope and the y-intercept (f'(x)=(-1/2)x - 1/2).
If the function is transformed by a vertical shift, the equation for the tangent line only changes slightly (f(x)=√(x) + 2). The slope is unaffected and there for stays equal to 1/2, however, the y-intercept has to be shifted the corresponding number of units up or down (f'(x)=(1/2)x + 5/2).
If the function undergoes a vertical stretch, the slope changes as well as the y-intercept (f(x)=2√(x)). When you graph this stretch, you can re-write the equation for the tangent line as: f'(x)=x+1. Because the slope has been vertically stretched by a factor of 2, you must signify this in the equation for the tangent line.
For the composition of functions, this works as well, but it is much more complex. In order to use this method to determine the equation of the tangent line of a composition of functions equation, you must follow the Chain Rule:
For; f o g(x) = f(g(x)), f o g'(x) = f'(g(x))(g'(x))
where you use the inside and outside of the function; you can refer to the inside as (u).
Ex: f(x)=(x^2 - 1)^4, u=x^2 - 1
First you can identify a point on the graph by plugging in the given x-value answers to the original equation for the y-value. If you're given f(x)=√(x^3+1) at x=2, you can plug in the 2: f(x)=√((2)^3+1) -------> f(x)=3 therefore, (2,3) is on the graph. Now you are required to find the tangent line, so you must follow the chain rule. First, this can be re-written as
f(x)=(x^3+1)^(1/2). So, in this problem, u=x^3+1 therefore, when you begin to write your derivative you can use (u) as opposed to the whole thing. Now, you can write f(u)=(u)^(1/2) making the derivative f'(u)=1/2u^(-1/2). Which you can further breakdown into f'(u)=(1)/(2√(u)). Then, find f'(u) at x=2: u=(2)^3+1 --------> u=9. The slope of the tangent line becomes: f'(9)=(1)/(2√(9)) -------> f'(9)=1/6. So, y=(1/6)x + 3/2. Now, identify the derivative of (u) at x=2, and the equation of that tangent line:
f(x)=x^3+1 -------> f'(x)=3x^2 -------> f'(2)=3(2)^2 -------> f'(2)=12
Equation of the tangent line: y=12x - 15
Following the chain rule, re-write the slope for the tangent line with the slope (1/6) plugged in for m in the point-slope formula, and the tangent line equation for (u) in for the x. You will end up with:
y=1/6(12x-15) + 3/2
This link is a graph of the example displaying the use of the chain rule to determine the equation for the tangent line of a composition of functions equation.
f(x)=√(x) -------> f(x)=x^(1/2)
Therefore, when derived for the function of the tangent line using the power rule, the equation becomes:
f'(x) = (1/2)x + b
After finding the y-intercept to be 1/2, you now have the equation of the tangent line:
f(x) = (1/2)x + 1/2
This is true for any other specific root of a function as well. If the function is transformed by being reflected over the x-axis (f(x)=-√(x)), the only thing that changes in the equation for the tangent line is the sign of the slope and the y-intercept (f'(x)=(-1/2)x - 1/2).
If the function is transformed by a vertical shift, the equation for the tangent line only changes slightly (f(x)=√(x) + 2). The slope is unaffected and there for stays equal to 1/2, however, the y-intercept has to be shifted the corresponding number of units up or down (f'(x)=(1/2)x + 5/2).
If the function undergoes a vertical stretch, the slope changes as well as the y-intercept (f(x)=2√(x)). When you graph this stretch, you can re-write the equation for the tangent line as: f'(x)=x+1. Because the slope has been vertically stretched by a factor of 2, you must signify this in the equation for the tangent line.
For the composition of functions, this works as well, but it is much more complex. In order to use this method to determine the equation of the tangent line of a composition of functions equation, you must follow the Chain Rule:
For; f o g(x) = f(g(x)), f o g'(x) = f'(g(x))(g'(x))
where you use the inside and outside of the function; you can refer to the inside as (u).
Ex: f(x)=(x^2 - 1)^4, u=x^2 - 1
First you can identify a point on the graph by plugging in the given x-value answers to the original equation for the y-value. If you're given f(x)=√(x^3+1) at x=2, you can plug in the 2: f(x)=√((2)^3+1) -------> f(x)=3 therefore, (2,3) is on the graph. Now you are required to find the tangent line, so you must follow the chain rule. First, this can be re-written as
f(x)=(x^3+1)^(1/2). So, in this problem, u=x^3+1 therefore, when you begin to write your derivative you can use (u) as opposed to the whole thing. Now, you can write f(u)=(u)^(1/2) making the derivative f'(u)=1/2u^(-1/2). Which you can further breakdown into f'(u)=(1)/(2√(u)). Then, find f'(u) at x=2: u=(2)^3+1 --------> u=9. The slope of the tangent line becomes: f'(9)=(1)/(2√(9)) -------> f'(9)=1/6. So, y=(1/6)x + 3/2. Now, identify the derivative of (u) at x=2, and the equation of that tangent line:
f(x)=x^3+1 -------> f'(x)=3x^2 -------> f'(2)=3(2)^2 -------> f'(2)=12
Equation of the tangent line: y=12x - 15
Following the chain rule, re-write the slope for the tangent line with the slope (1/6) plugged in for m in the point-slope formula, and the tangent line equation for (u) in for the x. You will end up with:
y=1/6(12x-15) + 3/2
This link is a graph of the example displaying the use of the chain rule to determine the equation for the tangent line of a composition of functions equation.