Uses Of Differentiation | Summary & Examples

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1 Summary

1.1 Increasing Function

1.2 Decreasing Function

1.3 Example #1

1.4 Stationary Points

1.5 Example #2

1.6 There exist 3 types of stationary points:

1.7 Example #3

Summary

If y = f(x) is an increasing function then $\frac { dy }{ dx } \quad >\quad 0$
If y = f(x) is a decreasing function then $\frac { dy }{ dx } \quad <\quad 0$
Stationary Points are obtained by solving the equation $\frac { dy }{ dx } \quad =\quad 0$
3 types of stationary points:

i) Maximum Point

ii) Minimum Point

iii) Point of Inflexion

Increasing Function

Refer to Fig 1, consider a point A on the curve y = f(x). A tangent is drawn at A and the gradient of the tangent at A is $\frac { dy }{ dx }$ .

In this situation when x increases the function f(x) also increases so $\frac { dy }{ dx }$ positive. Hence, it is known as an increasing function.

If y = f(x) is an increasing function then $\frac { dy }{ dx } \quad >\quad 0$ .

Decreasing Function

Refer to Fig 2, A tangent is drawn at A and the gradient of the tangent at A is $\frac { dy }{ dx }$ .

In this situation when x increases the function f(x) decreases so $\frac { dy }{ dx }$ is negative. Hence, it is known as a decreasing function.

If y = f(x) is a decreasing function then $\frac { dy }{ dx } \quad <\quad 0$ .

Example #1

Q. Find the range of values of x such that $f\left( x \right) \quad =\quad 3{ x }^{ 2 }\quad +\quad 4x\quad -\quad 7$ is an increasing function.

Solution:

$y\quad =\quad 3{ x }^{ 2 }\quad +\quad 4x\quad -\quad 7$

$\frac { dy }{ dx } \quad =\quad 6x\quad +\quad 4$

As f(x) is an increasing function hence $\frac { dy }{ dx } \quad >\quad 0$ .

So,

$6x\quad +\quad 4\quad >\quad 0$

$x\quad >\quad \frac { -2 }{ 3 }$ Ans

Stationary Points

Refer to Fig 3, consider a graph of y = f(x). Notice the 4 main situations in the diagram.

i) Along AB, gradient of the curve is increasing, therefore $\frac { dy }{ dx } \quad >\quad 0$ .
ii) On reaching point B, the tangent at B becomes parallel to the x axis. Hence, at B $\frac { dy }{ dx } \quad =\quad 0$ .
iii) After passing B, the gradient of the curve becomes negative, therefore $\frac { dy }{ dx } \quad <\quad 0$ .
iv) As the curve now reaches C, $\frac { dy }{ dx }$ at C is again 0.

To conclude, point B and C are special points where the tangents being parallel to the x axis, $\frac { dy }{ dx }$ is 0.

Stationary Points are obtained by solving the equation $\frac { dy }{ dx } \quad =\quad 0$

Example #2

Q. Find the stationary values of $\quad f\left( x \right) \quad =\quad 4{ x }^{ 3 }\quad -\quad 3x\quad +\quad 12$

Solution:

$\quad f\left( x \right) \quad =\quad 4{ x }^{ 3 }\quad -\quad 3x\quad +\quad 12$

Find $\frac { dy }{ dx }$

$\frac { dy }{ dx } \quad =\quad f^{ ' }\left( x \right) \quad =\quad 12{ x }^{ 2 }\quad -\quad 3$

We know that at stationary points $\frac { dy }{ dx } \quad =\quad 0$ .

Hence:

$12{ x }^{ 2 }\quad -\quad 3\quad =\quad 0$

${ x }^{ 2 }\quad =\quad \frac { 1 }{ 4 }$

$x\quad =\quad \pm \frac { 1 }{ 2 }$

By substituting both values of x at $\frac { dy }{ dx } \quad =\quad 0$ in f(x), we can find the stationary points.

At $x\quad =\quad +\frac { 1 }{ 2 }$

$f\left( \frac { 1 }{ 2 } \right) \quad =\quad 4{ (\frac { 1 }{ 2 } ) }^{ 3 }\quad -\quad 3(\frac { 1 }{ 2 } )\quad +\quad 12$

$=\quad 11$

At $x\quad =\quad -\frac { 1 }{ 2 }$

$f\left( -\frac { 1 }{ 2 } \right) \quad =\quad 4{ (-\frac { 1 }{ 2 } ) }^{ 3 }\quad -\quad 3(-\frac { 1 }{ 2 } )\quad +\quad 12$

$=\quad 13$

Ans: Therefore coordinates of the stationary points are $(\frac { 1 }{ 2 } ,\quad 11)$ & $(-\frac { 1 }{ 2 } ,\quad 13)$

A stationary point is also called a ‘turning point’ if it is either a maximum point or minimum point.

We are now aware that a curve has a stationary point where $\frac { dy }{ dx } \quad =\quad 0$ .

There exist 3 types of stationary points:

1. Maximum Point

In Fig 4 , before the stationary point (A), the gradient $\frac { dy }{ dx }$ is positive and immediately after the stationary point (A), $\frac { dy }{ dx }$ is negative . Hence, A is a maximum point.

2. Minimum Point

In Fig 5 , before the stationary point (A), the gradient $\frac { dy }{ dx }$ is negative and immediately after the stationary point (A), $\frac { dy }{ dx }$ is positive . Hence, A is a minimum point.

3. Point of Inflexion

In Fig 6, points A and B are neither maximum nor minimum. Although $\frac { dy }{ dx } \quad =\quad 0$ at both the points, but it does not change sign when moving through A and B i.e if $\frac { dy }{ dx }$ x is positive before A, it will remain positive after A as well. Similarly if $\frac { dy }{ dx }$ x is negative before B, it will remain negative after B as well. Such points are called ‘Point of inflexion’.

To find the nature of a stationary point, we need to find the second derivative of the function and substitute the stationary points in the equation. If our answer is positive it means that its a minimum point and vice versa.

If we differentiate a function again after finding $\frac { dy }{ dx }$ , it is called a second derivative $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ .

Remember:

$\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad \neq \quad { (\frac { dy }{ dx } ) }^{ 2 }\quad or\quad \quad \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \quad \neq \quad { (\frac { dy }{ dx } ) }^{ 3 }$

Example #3

Q. The equation of a curve is $y\quad =\quad { (2x\quad -\quad 3) }^{ 2 }\quad -\quad 6x$

i) Express $\frac { dy }{ dx }$ and $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ in terms of x.

ii) Find the coordinates of the two stationary points and determine the nature of each stationary point.

Solution:

i) $y\quad =\quad { (2x\quad -\quad 3) }^{ 2 }\quad -\quad 6x$

$\frac { dy }{ dx } \quad =\quad { 3{ (2x\quad -\quad 3) }^{ 2 } }(2)\quad -\quad 6$

$\frac { dy }{ dx } \quad =\quad 6{ (2x\quad -\quad 3) }^{ 2 }\quad -\quad 6\quad \quad \quad \quad \quad (1)$

$\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\quad 6\left[ 2(2x\quad -\quad 3)(2) \right] \quad -\quad 0$

$\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\quad 24(2x\quad -\quad 3)\quad \quad \quad \quad \quad (2)$

ii) Equate $\frac { dy }{ dx }$ to 0

$6{ (2x\quad -\quad 3) }^{ 2 }\quad -\quad 6\quad =\quad 0$

${ (2x\quad -\quad 3) }^{ 2 }\quad =\quad 1$

${ 2x\quad -\quad 3 }\quad =\quad \pm \quad 1$

Hence: $x\quad =\quad 2,\quad 1$

When: x = 2

$\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\quad 24(2x\quad -\quad 3)$

$=\quad 24(2(2)\quad -\quad 3)$

$=\quad 24$

Since $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad >\quad 0$ , x = 2 is a minimum point.

When: x = 1

$\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\quad 24(2x\quad -\quad 3)$

$=\quad 24(2(1)\quad -\quad 3)$

$=\quad -24$

Since $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad <\quad 0$ , x = 1 is a maximum point.