Differentiation allows us to find rates of change. For example, it allows us to find the rate of change of velocity with respect to time (which is acceleration). It also allows us to find the rate of change of x with respect to y, which on a graph of y against x is the gradient of the curve. There are a number of simple rules which can be used to allow us to differentiate many functions easily.
If y = some function of x (in other words if y is equal to an expression containing numbers and x’s), then the derivative of y (with respect to x) is written dy/dx, pronounced “dee y by dee x” .
Derivatives are defined as the varying rate of change of a function with respect to an independent variable. The derivative is primarily used when there is some varying quantity, and the rate of change is not constant
a function is denoted as y=f(x), the derivative is indicated by the following notations.
- D(y) or D[f(x)] is called Euler’s notation.
- dy/dx is called Leibniz’s notation.
- F’(x) is called Lagrange’s notation.
Differentiation Formulas List
Rolle’s theorem is one of the foundational theorems in differential calculus. It is a special case of, and in fact is equivalent to, the mean value theorem , which in turn is an essential ingredient in the proof of the fundamental theorem of calculus.
Suppose f(x) be a function satisfying three conditions:
1) f(x) is continuous in the closed interval a ≤ x ≤ b
2) f(x) is differentiable in the open interval a < x < b
3) f(a) = f(b)
Then according to Rolle’s Theorem, there exists at least one point ‘c’ in the open interval (a, b) such that:
f ‘ (c) = 0
A graphical demonstration of this will help our understanding; actually, you’ll feel that it’s very apparent:
In the figure above, we can set any two points as \(a,f(a)) and \(b,f(b)) as long as f(a)=f(b) and the function is differentiable within the interval (a,b). Then, of course, there has to be a point in between where f’=0,which is the red point in the diagram. Now let’s take a look at the mathematical proof of this theorem.
We divide it into two cases:
(1) f(x) is a constant function.
If f(x)f(x) is a constant function, then f’=0 for the whole interval. Then, of course, there exists a c such that f′(c)=0 within the interval (a,b).
(2) f(x) is not a constant function.
When f(x) is not a constant function but is continuous within the interval [a,b],according to the extreme value theorem f(x) must have a maximum function value and minimum function value within the interval [a,b].Since f(x)is not a constant function, at least one of the extrema must exist within the interval (a,b).
Expansion of function
Maxima and Minima from Calculus
One of the great powers of calculus is in the determination of the maximum or minimum value of a function. Take f(x) to be a function of x. Then the value of x for which the derivative of f(x) with respect to x is equal to zero corresponds to a maximum, a minimum or an inflexion point of the function f(x).
example, the height of a projectile that is fired straight up is given by the motion equation
Taking y0 = 0, a graph of the height y(t) is shown below
The derivative of a function can be geometrically interpreted as the slope of the curve of the mathematical function y(t) plotted as a function of t. The derivative is positive when a function is increasing toward a maximum, zero (horizontal) at the maximum, and negative just after the maximum. The second derivative is the rate of change of the derivative, and it is negative for the process described above since the first derivative (slope) is always getting smaller. The second derivative is always negative for a “hump” in the function, corresponding to a maximum.
Example: Find the maxima and minima for:
y = 5x3 + 2x2 − 3x
The derivative (slope) is:
y = 15x2 + 4x − 3
Which is quadratic with zeros at:
- x = −3/5
- x = +1/3
Could they be maxima or minima? (Don’t look at the graph yet!)
The second derivative is y” = 30x + 4
At x = −3/5: y” = 30(−3/5) + 4 = −14 it is less than 0, so −3/5 is a local maximum
At x = +1/3: y” = 30(+1/3) + 4 = +14it is greater than 0, so +1/3 is a local Minimum
Now you can look at the graph.)
The following steps are taken in the process of curve sketching:
Find the domain of the function and determine the points of discontinuity (if any).
Determine the x− and y−intercepts of the function, if possible. To find the x−intercept, we set y=0 and solve the equation for x. Similarly, we set x=0 to find the y−intercept. Find the intervals where the function has a constant sign (f(x)>0 and f(x)<0).
Determine whether the function is even, odd, or neither, and check the periodicity of the function. If f(−x)=f(x) for all x in the domain, then f(x) is even and symmetric about the y−axis. If f(−x)=−f(x) for all x in the domain, then f(x) is odd and symmetric about the origin.
Find the vertical, horizontal and oblique (slant) asymptotes of the function.
5. Intervals of Increase and Decrease
Calculate the first derivative f′(x) and find the critical points of the function. (Remember that critical points are the points where the first derivative is zero or does not exist.) Determine the intervals where the function is increasing and decreasing using the First Derivative Test.
6. Local Maximum and Minimum
Use the First or Second Derivative Test to classify the critical points as local maximum or local minimum. Calculate the y−values of the local extrema points.
7. Concavity/Convexity and Points of Inflection
Using the Second Derivative Test, find the points of inflection (at which f′′(x)=0). Determine the intervals where the function is convex upward (f′′(x)<0) and convex downward (f′′(x)>0).
8 Graph of the Function
Sketch a graph of f(x) using all the information obtained above.
Successive differentiation & Liebnitz Theorem