Slope of the tangent line/derivative?
Also, what exactly is the required limit process when finding the derivative of something like f(x)=2x-1 and how do you use it?
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represents a straight line, and hence the expression ‘tangent line’ to the graph of f(x) is inappropriate. A tangent makes sense only in connection with a curved line.
When you differentiate the function f(x) from first principles, you evaluate:
lim h -> 0 ( f(x + h) – f(x) ) / h )
When f(x) = 2x – 1, this formula gives:
lim h->0 ( 2(x + h) – 1 – (2x – 1) ) / h
= lim h->0( 2h / h )
= 2.
The slope is constant.
This is what you expect for a straight line. It is just the gradient m as given in the ‘slope-intercept’ form
y = mx + c
of the equation:
y = 2x – 1.
If you had a quadratic function such as x^2 + 3x – 1, then the differentiation process would give 2x + 3.
That means the slope of a tangent to the graph at point x has a gradient 2x + 3.
If x = -1, for example, the gradient of the tangent at the point (-1, -3) is the value of 2x + 3 at that point, namely 1.
f ‘(x) = -2
Plug in x = -1
f ‘ (-1) = -2
So the slope of the tangent line at (-1,5) is -2.
To use limits to find a derivative, you calculate what is called the difference quotient.
lim [f(x + h) – f(x)]/[h]
h->0
So for f(x) = 2x – 1, f(x + h) = 2(x + h) – 1 = 2x +2h – 1
lim[2x + 2h – 1 – (2x – 1)]/h
h ->0
lim[2x + 2h – 1 -2x + 1]/h
h->0
lim[2h]/h
h->0
lim2
h->0
