Applications of Derivatives Test - 1

Concept:

Following steps to finding maxima and minima using derivatives.

Find the derivative of the function.

Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.

Now we have to find the second derivative.

• f"(x) is less than 0 then the given function is said to be maxima
• If f"(x) Is greater than 0 then the function is said to be minima

We know,

sin 2x = 2 sinx.cosx

Calculation:

Let, f(x) = sinx .cos x

= \(\rm\frac12\times sin2 x \)            (∵ \(\rm sin2 x =2sinx \ . cosx\)) 

f'(x) = \(\rm\frac12\times (2cos2 x)\)

= cos 2x

Now, f'(x) = 0 ⇒ cos 2x = 0

We know, \(\rm cos(\fracπ2)=0\)

∴ cos 2x = \(\rm cos(\fracπ2)\)

⇒ 2x = π/2 

⇒ x = π/4

Also, f''(x) = \(\rm-(4sin2 x)\)

⇒ f''(\(\fracπ4\)) = - 4 < 0

At x = \(\fracπ4\) , f(x) is maximum.

∴ f(\(\fracπ4\)) = \(\rm\frac12\times sin (\frac{π}{2})\) 

= 1/2

Hence, option (4) is correct.

Concept:

|-a| = a,  where a = constant

Calculation:

We have, f(x) = |x - 4|

At, x = 0, f(0) = |0 - 4| = 4

At, x = 2, f(2) = |2 - 4| = |-2| = 2

At, x = 4, f(4) = |4 - 4| = 0

At, x = -4, f(-4) = |-4 - 4| = 8

So , at x = 4, we get minimum value 

Hence, option (3) is correct.

Formula used:

\(\rm \frac{d}{dx} x^{n} = nx^{n - 1}\)

Concept Used:

If f''(a) > 0 then x = a is a point of local minima

If f''(a) < 0 then x = a is a point of local maxima

Calculation:

Let f(x) = x + \(\frac{1}{x}\)​

\(\rm \frac{d f(x)}{dx} = 1 - \frac{1}{x^{2}}\)

\(\rm \frac{d f(x)}{dx} = 0\)

x = ±1

\(\rm \frac{d^{2} f(x)}{dx^{2}} = \frac{2}{x^{3}}\)

At x = 1, \(\rm \frac{d^{2} f(x)}{dx^{2}} = 2\) 

\(\rm \frac{d^{2} f(x)}{dx^{2}} > 0\)

So, the function has minima and its value is f(1) = 2

At x = -1, \(\rm \frac{d^{2} f(x)}{dx^{2}} = -2\)

\(\rm \frac{d^{2} f(x)}{dx^{2}} < 0\)

So, the function has maxima and its value is f(-1) = -2

⇒ fmax < fmin 

∴ Statement 1 is only correct.

Concept:

Trigonometric formulas used:

• sin 3x = 3sin x – 4sin³ x

Calculation:

Given: f(x) = 3sin x – 4sin³ x

We know that, sin 3x = 3sin x – 4sin³ x

So, f(x) = 3 sin x – 4 sin³ x = sin 3x

We know that, sin 3x is a periodic function.

As we know Sin x is a periodic function [-π/2 to π/2]  

Since here the function is Sin 3x the range becomes [-π/6 to π/6]

So, sin 3x increases from –π/6 to π/6.

∴ The length of the increasing interval = π/6 – (-π/6) = π/3 

Concept:

Following steps to finding maxima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have find second derivative.
  • f"(x) is less than 0 then the given function is said to be maxima

Calculation:

Let f(x) = xy + 5,

2x + y = 4 

⇒ y = 4 - 2x

So, f(x) =  x(4 - 2x) + 5 

f(x) = 4x - 2x² + 5

Differentiating with respect to x, we get

f'(x) = 4 - 4x 

Set f(x) = 0 

⇒ 4 - 4x = 0

⇒ x = 1

Again differentiating with respect to x

f''(x) = -4 < 0

Hence function is said to be maxima

So, maximum value will obtain at x = 1, and y = 4  - 2(1) = 2

∴ The maximum value of xy + 5 = 1(2) + 5 = 7

Hence, option (4) is correct. 

Concept:

A function f(x) is maximum at a point, where,

f ' (x) = 0 and f " (x) < 0

Calculation:

Given, s = 64t - 16t2,

So, the function s(t) is maximum when, 

s'(t) = 0 and s"(t) < 0

⇒ s'(t) = 64 - 16(2t) = 0 

⇒ 64 = 32 t 

⇒ t = 2

And at t = 2, s"(t) = -32 < 0

So, s(t) is maximum at t = 2.

So, The time required to reach the maximum height by the stone is 2s.

∴ The correct answer is an option (2).

Concept:

Differentiation formula

Quotient rule:

\(\rm \frac{d}{dx} (\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}\)

Calculation:

f(x) = \(\rm \frac{(log x)}{x}\)

⇒ f^'(x) = \(\rm \frac{1}{x^{2}} - \frac{log x}{x^{2}}\)

⇒ f′(x) = 0 at x = e

⇒ f′(x) > 0 when x < e

⇒ f′(x) < 0 when x > e

So, f(x) is maxima at x = e

⇒  f(e) = \(\rm \frac{1}{e}\)

∴ The maximum value of \(\rm \frac{(log x)}{x}\) is \(\rm \frac{1}{e}\)

Concept:

 Decreasing rate always represent by negative sign and increasing rate by positive sign.

Area of rectangle is given by A = x × y where x is length and y is breadth

Calculations:

Here \(\frac{{dx}}{{dt}} = \; - 7\;and\;\frac{{dy}}{{dt\;}} = 6\) and x = 10, y = 8

Area of Rectangle A = xy

The rate of change of area of rectangle

\(\frac{{dA}}{{dt}} = x\frac{{dy}}{{dt}} + y\frac{{dx}}{{dt}}\)

\(\frac{{dA}}{{dt}} = \left( {10} \right)\left( 6 \right) + 8\left( { - 7} \right)\)  

\(\frac{{dA}}{{dt}} = 4\)

Concept:

Formula:

• cos-1 (cos x) = x
• sin^-1 (sin x) = x
• Slope of the curve = tan θ = \(\rm \frac {dy}{dx}\)

Calculation:

Given: Slope = tan θ

y = sin-1 (cos x)

⇒ y = sin-1 (sin (π/2 -x))

⇒ y = π/2 – x                                        (∵ sin-1 (sin x) = x)

Differentiating both sides with respect to x, we get

\(\rm \frac {dy}{dx}\) = -1

As we know, Slope = \(\rm \frac {dy}{dx}\)

⇒ tan θ  = -1

∴ θ = 3π/4

Concept:

• If f′(x) > 0 then the function is said to be increasing.
• If f′(x) < 0 then the function is said to be decreasing.

Calculation:

Given:

f(x) = x - e^x

Differentiating with respect to x, we get

⇒ f’(x) = 1 - e^x

For increasing function,

f'(x) > 0

⇒ 1 – e^x > 0

⇒ e^x < 1

⇒ e^x < e⁰

∴ x < 0

So, x ∈ (-∞, 0)