# Uses Of Differentiation

### 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.