__maths__>__Differential Calculus__>__Introduction to Differential Calculus__### Differentiation: Graphical Meaning

In this page, graphical meaning of differentiation is discussed with examples.

*click on the content to continue..*

Summary of differentiation: Given `y=f(x)` a function of variable `x`, the derivative of `y` is

`(dy)/(dx)``= lim_(delta->0) (f(x+delta) - f(x))/delta`.

Which of the following is correct?

- The derivative is another function of variable `x`
- The derivative is another function of variable `x`
- The derivative is a numerical value

The answer is "The derivative is another function of variable `x`".

For some functions, the derivatives are numerical values, but it is not a numerical value for all functions.

Let us consider `y=(x^2)/2` and `(dy)/(dx) = x`. The figure depicts both the functions. Blue color curve is `y=(x^2)/2` and orange color line is `(dy)/(dx) = x`. *note: The plot is not to the scale on x and y axes.*

How are these two plots related?

- the rate of change of blue curve is plotted as orange line
- the rate of change of blue curve is plotted as orange line
- the two plots are not related

The answer is "the rate of change of blue curve is plotted as orange line". *The rate of change is a function of `x`*

Let us consider `y=(x^2)/2` and `(dy)/(dx) = x` given in the figure. The figure zooms in a small part of the plots. What do the two values at `x=a` signify?

- `y(a)` is the value function evaluates to at `x=a`
- `y′(a)` is the rate of change of `y` at `x=a`
- both the above
- both the above

The answer is "both the above". This result is known from the algebraic derivations. Let us see what this means in the given curve.

Considering `y=(x^2)/2` and `(dy)/(dx) = x` given in the figure. The derivative in first principles is given as

`y′(x) `

`= d/(dx) y`

`= lim_(delta x->0) (delta y)/(delta x)`

`delta x` at `x=a` is shown in the figure.

`delta y` for the `delta x` is shown in the figure.

How is the derivative `y′` related to `delta y` and `delta x`?

- derivative is not related to `delta x`
- the derivative is derived using limit `delta x` tending to `0`
- the derivative is derived using limit `delta x` tending to `0`

The answer is "the derivative is derived using limit `delta x` tending to `0`"

Considering `y=(x^2)/2` and `(dy)/(dx) = x` given in the figure. The figure shows non-zero `delta x` and corresponding `delta y` before limit is applied.

The points `P` and `Q` are on the curve separated by `delta x` on `x=a`. Consider the line passing through the points `P` and `Q`. Is the line a secant or a tangent?

- secant
- secant
- tangent

The answer is "Secant", as the line passes through two points on the curve.

Considering `y=(x^2)/2` and `(dy)/(dx) = x` given in the figure. It shows non-zero `delta x` and corresponding `delta y` before limit is applied. What is the slope of the line `bar(PQ)`?

- `(delta y)/(delta x)`
- `(delta y)/(delta x)`
- slope cannot be found.

The answer is "`(delta y)/(delta x)`". The line passing through the points `P(x_1, y_1)` and `Q(x_1+delta x, y_1+delta y)`

Slope of the secant

`=(y_2-y_1)/(x_2-x_1)`

`=(delta y)/(delta x)`

Considering `y=(x^2)/2` and `(dy)/(dx) = x` given in the figure. The derivative in first principles is given as

`y′(x) `

`= lim_(delta x->0) (delta y)/(delta x)`

Applying the limit, the `delta x` and `delta y` reduces "close to `0`", which is shown in red.

The points `P` and `Q` at the two ends of `dx` moves towards each other. When limit `delta x` is close to zero, what happens to the two points `P` and `Q`?

- merge to become a single point
- merge to become a single point
- cross over and move away

The answer is "merge to become a single point". The points `P` and `Q` move towards each other and merge.

Considering `y=(x^2)/2` and `(dy)/(dx) = x` given in the figure. Applying the limit, the `delta x` and `delta y` reduces close to `0`, which is shown in red.

The points `P` and `Q` merge to a single point. The line passing through `P` and `Q` is shown in red. As the limit is applied, the line touches the curve in only one point

Is the line a secant or a tangent?

- secant
- tangent
- tangent

The answer is "tangent", as the line touches at a single point on the curve.

Considering `y=(x^2)/2` and `(dy)/(dx) = x` given in the figure. Applying the limit, the `delta x` and `delta y` reduces close to `0`, which is shown in red. What is the slope of the tangent at `x=a`?

- `(dy)/(dx)|_(x=a)`
- `(dy)/(dx)|_(x=a)`
- slope cannot be calculated

The answer is "`(dy)/(dx)|_(x=a)`". The rate of change is the slope at that point.

`(dy)/(dx)|_(x=a)` is the slope of the tangent on curve `y` at position `x=a`.

For a function `f(x)`, the slope of tangent at `x=a` is `d/(dx) f(x)|_(x=a)`. The figure shows `f(x)` in blue and `d/(dx) f(x)` in orange. Three positions are identified with `3` vertical lines. Which of the following describe the slope of `f(x)` at the position `P`?

- negative or decreasing rate of change
- negative or decreasing rate of change
- positive or increasing rate of change
- zero or no change for small change in `x`

The answer is "negative or decreasing rate of change". This is also evident from the negative value at the same position in orange line.

For a function `f(x)`, the slope of tangent at `x=a` is `d/(dx) f(x)|_(x=a)`. The figure shows `f(x)` in blue and `d/(dx) f(x)` in orange. Which of the following describe the slope of `f(x)` at the position `Q`?

- negative or decreasing rate of change
- positive or increasing rate of change
- positive or increasing rate of change
- zero or no change for small change in `x`

The answer is "positive or increasing rate of change". This is also evident from the positive value at the same position in orange line.

Figure shows `y=x+1` in blue and `(dy)/(dx) = 1` in orange. Which of the following is observed?

- the function has constant rate of change
- the function has constant rate of change
- the rate of change of the function is `0`

The answer is "the function has constant rate of change". This is evident from the flat orange line.

The figure shows `y=sin x` in blue and `(dy)/(dx) =cos x` in orange. Which of the following is observed?

- when `y` reaches maximum, the rate of change is `0`
- when `y` reaches minimum, the rate of change is `0`
- both the above
- both the above

The answer is "both the above". This is evident from the values on orange line, the derivative of the function, having value 0 at the positions of maxima and minima.

The figure shows `y` in blue and `(dy)/(dx)` in orange. Two positions are identified. Which of the following is observed?

- when `y` reaches maximum, the rate of change crosses `0` from positive to negative
- when `y` reaches minimum, the rate of change crosses `0` from negative to positive
- both the above
- both the above

The answer is "both the above". This is evident from the values on orange line, the derivative of the function, crossing the x axis at the positions of maxima and minima.

The slope of the tangent on curve is the derivative evaluated at that point.

**Graphical Meaning of Derivative:** For a function `f(x)`, the derivative `f′(x)` is another function of the variable. • at a point `x=a`, the rate of change of the curve is `f′(a)`

• at a point `x=a`, the slope of the tangent is `f′(a)`.

• at the maxima of the curve `f(a_1)`, the derivative `f′(a_1)` crosses `0` from positive rate of change to negative rate of change.

• at the minima of the curve `f(a_2)`, the derivative `f′(a_2)` crosses `0` from negative rate of change to positive rate of change.

*Solved Exercise Problem: *

The figure shows `y` in blue and `(dy)/(dx)` in orange. Which of the following observation is true for the given function?

- for negative values of `x`, the rate of change is negative
- for positive values of `x`, the rate of change is positive
- at `x=0`, the rate of change crosses `0` from negative to positive
- all the above
- all the above

The answer is "all the above"

*slide-show version coming soon*