use the definition of the derivative to calculate f'(x), if f(x)= SQRT(2x) and find the equations of the tangent line and the normal line to the graph of f at x=2
f(x) = √(2x)
f '(x) = (/2)(2)(2x)^(-1/2) or 1/√(2x)
f(2) = 2
f'(2) = 1/2
tangent equation:
x - 2y = c
at (2,2)
2-4=c
c = -2
tanglent: x-2y=-2
normal: slope = -2
equation:
2x + y = k
at (2,2,)
4+2 = k = 6
normal : 2x + y = 6
To calculate the derivative of f(x) using the definition of the derivative, we need to apply the limit definition. The definition of the derivative of a function f(x) at a given point x = a is:
f'(a) = lim(h→0) [ f(a + h) - f(a) ] / h
Now, let's find the derivative of f(x) step by step:
1. Start with the given function:
f(x) = √(2x)
2. Substitute the value of a into the function:
f(a) = √(2a)
3. Expand the function with the value of a + h:
f(a + h) = √(2(a + h))
4. Apply the definition of the derivative:
f'(a) = lim(h→0) [ √(2(a + h)) - √(2a) ] / h
5. Rationalize the numerator:
f'(a) = lim(h→0) [ (√(2(a + h)) - √(2a)) * (√(2(a + h)) + √(2a)) ] / (h * (√(2(a + h)) + √(2a)))
6. Simplify the numerator:
f'(a) = lim(h→0) [ 2(a + h) - 2a ] / (h * (√(2(a + h)) + √(2a)))
7. Simplify further:
f'(a) = lim(h→0) [ 2h ] / (h * (√(2(a + h)) + √(2a)))
8. Cancel the common factors:
f'(a) = lim(h→0) [ 2 ] / (√(2(a + h)) + √(2a))
9. Substitute the value of a = 2 into the expression:
f'(2) = 2 / (√(2(2 + h)) + √(2(2)))
10. Simplify further:
f'(2) = 2 / (√(4 + 2h) + √(4))
11. Simplify the expression under the square root:
f'(2) = 2 / (√(2h + 4) + 2)
Now, to find the equations of the tangent and normal lines at x = 2, we need to use the derivative f'(x) at that point.
At x = 2, f'(2) = 2 / (√(2h + 4) + 2)
1. Equation of the tangent line:
The slope of the tangent line is the value of the derivative at x = 2, which is f'(2).
Let's assume the equation of the tangent line is y = mx + c, where m is the slope and c is the y-intercept.
Since the slope of the tangent line is f'(2), the equation of the tangent line is:
y = (f'(2) * (x - 2)) + f(2)
Plug in f'(2) and f(2) into the equation to get the final form of the tangent line.
2. Equation of the normal line:
The slope of the normal line is the negative reciprocal of the slope of the tangent line at x = 2.
Let's assume the equation of the normal line is y = mx + c, where m is the slope and c is the y-intercept.
The slope of the normal line is -1/f'(2).
Therefore, the equation of the normal line is:
y = (-1/f'(2) * (x - 2)) + f(2)
Substitute the values of f'(2) and f(2) into the equation to obtain the final form of the normal line equation at x = 2.