Introduction

The subject of calculus can broadly be defined as the study of changes. The differential calculus is one major part of the subject. In differential calculus we study about the changes at infinitesimal stage. In other words, differential calculus is microscopic study of changes.

The derivative of a function is the instantaneous rate of change of function at any point. We can use the table of derivatives for standard functions while finding the derivatives for harder functions. Also certain rules of differentiation which help us while differentiating functions that are complex in nature.

There are a number of applications of derivatives like finding maximum and minimum of functions, optimizing functions, equations of tangents and normal at the desired points, finding rate of changes, evaluating approximate values etc.

Let us take a short but closer look of important concepts of differential calculus.

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The subject of calculus can broadly be defined as the study of changes. The differential calculus is one major part of the subject. In differential calculus we study about the changes at infinitesimal stage. In other words, differential calculus is microscopic study of changes.

The derivative of a function is the instantaneous rate of change of function at any point. We can use the table of derivatives for standard functions while finding the derivatives for harder functions. Also certain rules of differentiation which help us while differentiating functions that are complex in nature.

There are a number of applications of derivatives like finding maximum and minimum of functions, optimizing functions, equations of tangents and normal at the desired points, finding rate of changes, evaluating approximate values etc.

Let us take a short but closer look of important concepts of differential calculus.

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## Calculus Derivative Examples

The differentiation of this function is to be done in various steps, using different rules of differentiation and using the table of standard derivatives. First let us use the quotient rule for the whole function. That is,

f '(x) = $\frac{\frac{d(arc cos(x/2))}{dx}*\sqrt{2x + 7}- \frac{d(\sqrt{2x + 7})}{dx}*arccos(x/2)}{(\sqrt{2x + 7})^2}$

^{}Now, let us do part by part.

$\frac{d(arc cos(x/2))}{dx}= \frac{-1}{\sqrt{1 - (x^2/4)}}$*(1/2) = $\frac{-1}{\sqrt{4 - x^2}}$, treating (x/2) as a separate function and using chain rule.

$\frac{d(\sqrt{2x + 7})}{dx}$ = (1/2)*$\frac{1}{\sqrt{2x + 7}}$*(2) = $\frac{1}{\sqrt{2x + 7}}$, treating (2x + 7) as a separate function,using chain rule and power rule for square root.

Therefore, f '(x) = -[$\frac{1 }{\sqrt{4 - x^2}\sqrt{2x + 7}}$+$\frac{arc cos(x/2)}{(2x + 7)\sqrt{2x + 7}}$]

Example 2: Find the derivative of the function f(x) = {ln (x)}

^{ln (x)}

The differentiation here is done by the method called 'logarithmic differentiation'. That is explained as below. Taking natural logarithm on both sides,

ln [f(x)] = ln [{ln (x)}

^{ ln (x)}] = ln (x) * ln [{ln (x)}

Differentiate both sides using product rule. Also use chain rule on the left side and for the second term on the right. That is,

$\frac{1}{f(x)}$*f '(x) = $\frac{ln[ln(x)]}{x}+\frac{1}{x}$

f '(x) = f(x)*[$\frac{ln[ln(x)]}{x}+\frac{1}{x}$] = $\frac{f(x)}{x}$[ln[ln(x)]+ 1] or,

f '(x) = $\frac{ln (x)^{ln(x)}}{x}$[ln[ln(x)]+ 1]

Example 3: A man 6 ft tall, is walking away at a speed of 3 mph from an illuminated lamp post of 18 ft high. What is the rate at which his shadow will be increasing?

The following diagram explains the situation described in the problem.

AB is the lamp post and AB is the position of the man at any time. DE is the shadow of the man formed at that time. It can be seen that the triangles AEB and CED are similar. Equating the ratios of corresponding parts of similar triangle,

DE/BE = CD/AB or y(/x + y) = h/H. Since H = 18 ft and h = 6 ft,

y/(x + y) = 6ft/18ft = 1/3 or, 3y = x + y or y = (1/2)x

Differentiating both sides with respect to 't', dy/dt = (1/2)(dx/dt)

dx/dt is the instantaneous rate of change the distance between the lamp post and the man = speed at which the man is walking = 3 mph.

Therefore, dy/dt = (1/2)(3 mph) = 1.5 mph.

Hence the length of shadow is increasing at the rate of 1.5 mph.

Example 4: A car starts from a point with a speed of 30 mph towards north and another car starts form the same point at the same time but towards north with a speed of 40 mph. Find the rate of change of the distance between the cars at two hours later.

Let both the cars start from the point O and let A and B be the positions of the cars at any time. The distance AB between the cars at any time can be determined by Pythagorean theorem. That is,

AB

^{2}= OA

^{2}+ OB

^{2}or, s

^{2}= x

^{2}+ y

^{2}. Differentiating both sides with respect to time ‘t’, 2s (ds/dt) = 2x (dx/dt) + 2y (dy/dt) or, s (ds/dt) = x (dx/dt) + y (dy/dt)

After 2 hours, the car A is at 2*30 = 60 miles from O and the car B is at 2*40 = 80 miles from O. And the distance between them is, √(60

^{2}+ 80

^{2}) = √(3600 + 6400) = √(10000) = 100 miles. Therefore, after 2 hours,

100 (ds/dt) = 60 (dx/dt) + 80 (dy/dt). The derivatives (dx/dt) and (dy/dt) are nothing but the instantaneous speeds of cars which are given as 30 mph and 40 mph respectively. Therefore,

100 (ds/dt) = 60 (30) + 80 (40) = 100 (ds/dt) = 1800 + 3200 = 5000 or, ds/dt = 50 mph. Therefore, the rate of change of distance between the two cars after two hours is 50 mph.

Example 5: For the function f(x) = 2x

^{3}-3x

^{2}-72x+12, find the followings.

1) The intervals of increasing and decreasing

2) The points of maximum and minimum

3) Point of inflection

4) Intervals of concave up and concave down

Let us first find the derivative of the given function.

f '(x) = 6x

^{2}- 6x - 72 = 6(x

^{2}- x - 12) = 6(x - 4)(x + 3)

The function is increasing in an interval where the derivative of the function is positive. That is when,

(x - 4)(x + 3) > 0

The above inequality is true when x > 4 and when x > -3. Both the conditions are satisfied only when x > 4.

It is also true when x < 4 and x < -3. Both the conditions are satisfied only when x < -3

Thus the intervals at which the derivative is positive are (-∞, -3) U (4, ∞).

Therefore, the intervals of the function increasing are (-∞, -3) U (4, ∞).

The function is decreasing in an interval where the derivative of the function is negative. That is when,

(x - 4)(x + 3) < 0

The above inequality is true either when x < 4 and x > -3 or when x > 4 and x < -3. Clearly, the second set of condition is impossible.

Thus, the interval in which the derivative becomes negative is (-3, 4).

Therefore, the interval of the function decreasing is (-3, 4).

It may be noted that at x = -3 and at x = 4, the derivatives become 0. But, on both sides of the neighborhood of x = -3, the function changes from increasing to decreasing and hence x = -3 is the point of maximum. Similarly, on both sides of the neighborhood of x = 4, the function changes from decreasing to increasing and hence x = 4 is the point of minimum. This can also be confirmed by second derivative test.

f '(x) = 6x

^{2}- 6x - 72 and therefore, f ''(x) = 12x - 6

At x = -3, f "(x) = 12(-3) - 6 = -42. Since it is negative, x = -3 is a point of maximum.

At x = 4, f "(x) = 12(4) - 6 = 42. Since it is positive, x = 4 is a point of minimum.

The second derivative f "(x) = 12 - 6. It is 0 when x =0.5. And f(0.5) = -24.5 Therefore, (0.5, -24.5) is the point of inflection.

The second derivative is negative in the interval (-∞, 0.5) and positive in the interval (0.5, ∞)

Therefore, the function is concave down in the interval (-∞, 0.5) and concave up in the interval (0.5, ∞)

Example 6: A open box is made from a rectangular tin sheet of 24in. x 45 in. by cutting squares of same size at the corners and bending up the sheet along the corners. Optimize the size of the square for maximum volume of the box.

The following diagram explains how the box is made.

Let 'x cm' be the sides of the squares that are cut at the corners. The sheet becomes a box when it is folded up along the dotted lines shown and the dimensions of the box may be seen l= (45 - 2x), w = (24 - 2x) and h = x

The volume of the box, V = l*w*h = (45 - 2x)*(24 - 2x)* x = 1080x - 138x

^{2}+ 4w

^{3}

For the volume to be maximum or minimum, V'(x) = 0.

That is, 1080 - 276x + 12x

^{2}= 0 or 12(x - 18)(x - 5) = 0 or x = 18 and x = 5

The second derivative V''(x) = -276+ 24w = -12(23 - 2w). It is negative when x = 5 and positive when x = 18. Therefore, for V to be maximum, x = 5.

Hence, for maximum value of the box, the sides of the squares to be cut are 5 in.

Example 7: Find the equations of the tangent and normal to the circle x2 + y2 = 25 at x = 4.

The given equation of the circle is x

^{2}+ y

^{2}= 25 or y

^{2}= 25 - x

^{2}or y = ±√ (25 - x

^{2})

Differentiating both sides with respect to 'x', dy.dx =± (1/2)[1/√ (25 - x

^{2})](-2x) = ± [-x/√ (25 - x

^{2})]

The slope of the tangent at x = 4 is the value of the derivative for x = 4. Since there are two roots, there are two tangents with opposite slopes. In other words, if m

_{1}and m

_{2}are the slopes of the tangents,

m

_{1}= - 4/√ (25 - 42) = 4/3 and m

_{2}= -[- 4/√ (25 - 42)] = 4/3

When x = 4, y = ±√ (25 - 42) = ± 3

Therefore, the points of tangency are (4, 3) and (4, -3). That is, these points lie on the tangents also. It may be noted that the tangent passing through (4, 3) will have a slope of -4/3 and the tangent passing through (4, -3) will have a slope of 4/3 The equations of the tangents can be found by using slope point formula,

y - 3 = (-4/3)(x - 4) or y - 3 = -(4/3)x + (16/3) or y = -(4/3)x + (25/3)

y + 3 = (4/3)(x - 4) or y+ 3 = (4/3)x - (16/3) or y = (4/3)x - (25/3)

Since a tangent and normal are perpendicular to each other, we can use the condition of perpendicularity to find the slope of a normal. The slope of the normal at (4, 3) is -1/(-4/3) = 3/4 and the slope of the normal at (4, -3) is -1/(4/3) = -3/4. Using again the slope point formula, the equations of normals are, y - 3 = (3/4)(x - 4) or y - 3 = (3/4)x - 3 or y = (3/4)x

y + 3 = (-3/4)(x - 4) or y+ 3 = (-3/4)x + 3 or y = (-3/4)x

Example 8: Find the approximate value of square root of 25.7

Let us assume a function y = f(x). Then the derivative dy/dx is defined as the limit of [f(x + Δx)- f(x)]/Δx or Δy/Δx, when Δx-> 0. For smaller intervals (that is when Δx is very small) dy can be approximately equal to Δy. and hence, Δy ≈ (dy/dx)Δx. We will use this concept to solve this problem.

Let y = √x dy/dx = (1/2)(1/√x)

In the given problem, 25. 7 can be written in the form (x + Δx) and hence x = 25 and Δx = 0.7

Therefore, Δy = (dy/dx)(Δx) = (1/2)(1/5)(0.7), since √x = √25 = 5

or, Δy = 0.7/10 = 0.07

Hence, √25.7≈ y + Δy ≈ 5 + 0.07 ≈ 5.07

Thus the approximate value of square root of 25.7 is 5.07.

(The accurate value for the same from a calculator is 5.0695167422546303294105151803584....!)