Use the limit definition of the derivative to find f′(x) if
1. f(x) = x^2 + 3x
2. f(x) = 1/x + 1
now what i am confuse with is that there is no value given for x. So how is one suppose to do these problems. please show steps and this is just the 3rd chapter of my calc book so don't use shortcuts that come in the future.
are you using the defn. of
LIm (f(x+h) - f(x)/h, as h ---> 0 ?
this will help you:
tutorial(dot)math(dot)lamar(dot)edu/Classes/CalcI/DefnOfDerivative.aspx
To find the derivative using the limit definition, we need to evaluate the following limit:
1. For f(x) = x^2 + 3x:
The derivative of f(x) is denoted as f'(x) or dy/dx.
We can use the limit definition of the derivative:
f'(x) = lim(h→0) [(f(x + h) - f(x))/h]
Let's substitute f(x) = x^2 + 3x into the formula:
f'(x) = lim(h→0) [((x + h)^2 + 3(x + h) - (x^2 + 3x))/h]
Now, expand and simplify the expression inside the limit:
f'(x) = lim(h→0) [(x^2 + 2xh + h^2 + 3x + 3h - x^2 - 3x)/h]
f'(x) = lim(h→0) [(2xh + h^2 + 3h)/h]
Next, cancel out the 'h' factor:
f'(x) = lim(h→0) [2x + h + 3]
Finally, take the limit as h approaches 0:
f'(x) = 2x + 3
2. For f(x) = 1/x + 1:
Apply the limit definition of the derivative again:
f'(x) = lim(h→0) [(f(x + h) - f(x))/h]
Substitute f(x) = 1/x + 1:
f'(x) = lim(h→0) [((1/(x + h) + 1) - (1/x + 1))/h]
Simplify the expression:
f'(x) = lim(h→0) [(1/(x + h) - 1/x)/h]
f'(x) = lim(h→0) [(x - (x + h))/(x(x + h)h)]
f'(x) = lim(h→0) [-h/(x(x + h)h)]
Cancel out the 'h' factor:
f'(x) = lim(h→0) [-1/(x(x + h))]
Take the limit as h approaches 0:
f'(x) = -1/(x(x + 0))
f'(x) = -1/(x^2)
Thus, the derivative of f(x) = x^2 + 3x is f'(x) = 2x + 3,
and the derivative of f(x) = 1/x + 1 is f'(x) = -1/(x^2).
To find the derivative using the limit definition, we need to consider the definition itself:
f'(x) = lim(h -> 0) [f(x + h) - f(x)] / h
Now let's apply this definition to each of the given functions:
1. f(x) = x^2 + 3x
First, we substitute x + h back into the function f(x) and subtract f(x) from it:
f(x + h) = (x + h)^2 + 3(x + h)
= x^2 + 2hx + h^2 + 3x + 3h
Next, we can subtract f(x) from f(x + h):
[f(x + h) - f(x)] = (x^2 + 2hx + h^2 + 3x + 3h) - (x^2 + 3x)
= 2hx + h^2 + 3h
Finally, divide by h and consider the limit as h approaches 0:
f'(x) = lim(h -> 0) [2hx + h^2 + 3h] / h
= lim(h -> 0) 2x + h + 3
= 2x + 3
Therefore, the derivative of f(x) = x^2 + 3x is f'(x) = 2x + 3.
2. f(x) = 1/x + 1
Similarly, we apply the limit definition to find the derivative:
First, substitute x + h back into the function f(x) and subtract f(x) from it:
f(x + h) = 1/(x + h) + 1
Now, subtract f(x) from f(x + h):
[f(x + h) - f(x)] = [1/(x + h) + 1] - [1/x + 1]
= 1/(x + h) - 1/x
To simplify, we need to find a common denominator:
[f(x + h) - f(x)] = (x - (x + h)) / [(x + h)x]
= -h / [(x + h)x]
Finally, divide by h and consider the limit as h approaches 0:
f'(x) = lim(h -> 0) [-h / [(x + h)x]] / h
= lim(h -> 0) -1 / [(x + h)x]
= -1 / (x^2)
Therefore, the derivative of f(x) = 1/x + 1 is f'(x) = -1 / (x^2).
Note: When using the limit definition of the derivative, you don't need a specific value of x. The definition gives you a general formula for finding the derivative at any point.