Applications of Derivatives in Maths & Real Life with Examples (2024)

Solved Examples of Applications of Derivatives

Well acknowledged with the various applications of derivatives, let us practice some solved examples to understand them with a mathematical approach.

Example 1: Find the rate of change of the area of a circle with respect to its radius r when r = 6 cm.

Solution:Here we have to find the rate of change of the area of a circle with respect to its radius r when r = 6 cm.

As we know, the area of a circle is given by: \(π ⋅ r^2\) where r is the radius of the circle.

Let A =\(π ⋅ r^2\)

Also, we know that, if y = f(x), then dy/dx denotes the rate of change of y with respect to x.

So, by differentiating A with respect to r we get,

⇒\(\frac{dA}{dr}=\frac{d}{dr}\left(\pi⋅r^2\right)=2\pi r\)

Now we have to find the value of dA/dr at r = 6 cm i.e \(\left[\frac{dA}{dr}\right]_{_{r=6}}\)

⇒\(\left[\frac{dA}{dr}\right]_{_{r=6}}=2\pi⋅6=12\pi\text{ cm }\)

Hence, the rate of change of the area of a circle with respect to its radius r when r = 6 cm is 12π cm.

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Example 2: Find the equation of a tangent to the curve \(y = x^4 – 6x^3 + 13x^2 – 10x + 5\) at the point (1, 3).

Solution: Given: Equation of curve is: \(y = x^4 – 6x^3 + 13x^2 – 10x + 5\).

Here we have to find the equation of a tangent to the given curve at the point (1, 3).

As we know the slope of the tangent at any point say \((x_1, y_1)\) to a curve is given by: \(m=\left[\frac{dy}{dx}\right]_{_{(x_1,y_1)}}\)

⇒\(\frac{dy}{dx}=4x^3−18x^2+26x−10\)

⇒\(m=\left[\frac{dy}{dx}\right]_{_{(1,3)}}=(4\times1^3−18\times1^2+26\times1−10)=2\)

So, the slope of the tangent to the given curve at (1, 3) is 2.

As we know the equation of tangent at any point say \((x_1, y_1)\) is given by: \(y−y_1=\left[\frac{dy}{dx}\right]_{_{(x_1,y_1)}}⋅(x−x_1)\)

Here, \(x_1 = 1, y_1 = 3\) and \(\left[\frac{dy}{dx}\right]_{_{(1,3)}}=2\)

Substituting these values in the equation:

⇒y−3=2⋅(x−1)

⇒ 2x – y + 1 = 0

Hence, the equation of the tangent to the given curve at the point (1, 3) is: 2x – y + 1 = 0

Learn derivatives of cos x, derivatives of sin x, derivatives of xsinx and derivative of 2x

Example 3: Amongst all the pairs of positive numbers with sum 24, find those whose product is maximum.

Solution:Let the pairs of positive numbers with sum 24 be: x and 24 – x.

Then let f(x) denotes the product of such pairs.

i.e \(f(x)=x\times(24-x)=24x-x^2\)

Here we have to find that pair of numbers for which f(x) is maximum.

First, we have to find f'(x)

⇒ f'(x) = 24 – 2x

Now let’s find the roots of equation f'(x) = 0

⇒ 2x = 24

⇒ x = 12

Now let’s find out f”(x) i.e \(\frac{d^2(f(x))}{dx^2}\)

⇒ f”(x) = – 2

Now evaluate the value of f”(x) at x = 12

⇒ f”(12) = – 2 < 0

As we know that according to the second derivative test if f”(c) < 0 then x = c is a point of maxima

So, x = 12 is a point of maxima

So, when x = 12 then 24 – x = 12

Hence, the required numbers are 12 and 12

Learn about Integral Calculus

Example 4: Find the Stationary point of the function \(f(x)=x^2−x+6\)

Solution:Given \(f(x)=x^2−x+6\)

As we know that point c from the domain of the function y = f(x) is called the stationary point of the function y = f(x) if f′(c)=0

First, find f′(x).

f’(x) = 2x – 1

Now,

f’(x) = 0

⇒ 2x – 1 =0

Stationary point of the function \(f(x)=x^2−x+6\) is 1/2.

Example 5: An edge of a variable cube is increasing at the rate of 5 cm/sec. How fast is the volume of the cube increasing when the edge is 10 cm long?

Concept:

• Volume of a cube is given by: a3, where a is the length of each side of the cube.
• If y = f(x), then dy/dx denotes the rate of change of y with respect to xits value at x = a is denoted by:\({\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}\)
• Decreasing rate is represented by negative sign whereas increasing rate is represented bypositive sign.
• Chain Rule:
• Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then\(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} \cdot \frac{{dv}}{{dx}}\)

At any instant t, let the length of each side of the cube be x, and V be its volume.

Given: dx/dt = 5 cm/sec

As we know that, volumeof a cube is given by: a3, where a is the length of each side of the cube.

i.e V = x³

Now by differentiating V with respect to t, we get

\(⇒ \frac{{dV}}{{dt}} = \frac{{dV}}{{dx}} \cdot \frac{{dx}}{{dt}}\)———(BY chain Rule)

\(⇒ \frac{{dV}}{{dx}} = \frac{{d\left( {{x^3}} \right)}}{{dx}} = 3{x^2}\)

Example 6: The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate 4 cm/minute. When x = 8 cm and y = 6 cm then find the rate of change of the area of the rectangle.

Concept;

• Area of rectangle is given by: a× b, where a is the length and b is the width of the rectangle.
• If y = f(x), then dy/dx denotes the rate of change of y with respect to xits value at x = a is denoted by:\({\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}\)
• Decreasing rate is represented by negative sign whereas increasing rate is represented bypositive sign.
• \(\frac{d}{{dx}}\left[ {f\left( x \right) \cdot g\left( x \right)} \right] = f\left( x \right) \cdot \;\frac{{d\left\{ {g\left( x \right)} \right\}}}{{dx}}\; + \;\;g\left( x \right) \cdot \;\frac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}}\)

​Calculation:

At any instant t, let A be the area of rectangle, x be the length of the rectangle and y be the width of the rectangle.

Given: dx/dt = – 5cm/minute and dy/dt = 4cm/minute.

As we know that, areaof rectangle is given by: a× b, where a is the length and b is the width of the rectangle.

i.e A = x× y

As we know that,\(\frac{d}{{dx}}\left[ {f\left( x \right) \cdot g\left( x \right)} \right] = f\left( x \right) \cdot \;\frac{{d\left\{ {g\left( x \right)} \right\}}}{{dx}}\; + \;\;g\left( x \right) \cdot \;\frac{{d\left\{ {f\left( x \right)} \right\}}}{{dx}}\)

Now by differentiating A with respect to t we get,

\(\Rightarrow \frac{{dA}}{{dt}} = \frac{{d\left( {x \times y} \right)}}{{dt}} = \frac{{dx}}{{dt}} \cdot y + x \cdot \frac{{dy}}{{dt}}\)

Now by substituting the value of dx/dt and dy/dt in the above equation we get,

\(\Rightarrow \frac{{dA}}{{dt}} = \left( { – \;5} \right) \cdot y + x \cdot 6\)

Now substitute x = 8 cm and y = 6 cm in the above equation we get,

\(\Rightarrow \frac{{dA}}{{dt}} = \left( { – \;5} \right) \cdot 6 + 8 \cdot 6 = 2\;c{m^2}/min\)

Hence, the area of rectangle is increasing at the rate2 cm2/minute

Example 7: A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area is increasing when its radius is 5 cm? (Takeπ = 3.14)

Concept;

• Surface area of a sphere is given by: 4πr2where r is the radius of the sphere.
• If y = f(x), then dy/dx denotes the rate of change of y with respect to xits value at x = a is denoted by:\({\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}\)
• Decreasing rate is represented by negative sign whereas increasing rate is represented bypositive sign.
• Chain Rule:
• Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then\(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} \cdot \frac{{dv}}{{dx}}\)

As we know that soap bubble is in the form of a sphere. At an instant t, let its radius be r and surface area be S.

Given: dr/dt = 0.02 cm/sec

As we know the surface area of a sphere is given by: 4πr2where r is the radius of the sphere.

i.e S = 4πr²

So, by differentiating S with respect to t we get

\(\Rightarrow \frac{{dS}}{{dt}} = \frac{{dS}}{{dr}} \cdot \frac{{dr}}{{dt}}\)

\(\Rightarrow \frac{{dS}}{{dr}} = \frac{{d\left( {4π {r^2}} \right)}}{{dr}} = 8π r\)

By substituting the value of dS/dr in dS/dt we get

\(\Rightarrow \frac{{dS}}{{dt}} = 8π r \cdot \frac{{dr}}{{dt}}\)

By substituting r = 5 cm,π = 3.14 and dr/dt = 0.02 cm/sec in the above equation we get

\(\Rightarrow {\left[ {\frac{{dS}}{{dt}}} \right]_{r = 5}} = \left( {8 \times 3.14 \times 5 \times 0.02} \right) = 2.512\;c{m^2}/sec\)

Example 8: A stone is dropped into a quite pond and the waves moves in circles. If the radius of the circular wave increases at the rate of 8 cm/sec, find the rate of increase in its area at the instant when its radius is 6 cm?

Concept;

• Area of circle is given by:π⋅ r2where r is the radius of the circle.
• If y = f(x), then dy/dx denotes the rate of change of y with respect to xits value at x = a is denoted by:\({\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}\)
• Decreasing rate is represented by negative sign whereas increasing rate is represented bypositive sign.
• Chain Rule:
• Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then\(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} \cdot \frac{{dv}}{{dx}}\)

Calculation:

When the stone is dropped in the quite pond the corresponding waves generated moves in circular form.

Given:\(\frac{{dr}}{{dt}} = 8cm/\sec \)

So, here we have to find therate of increase inthe area of the circular waves formed at the instant when the radius r = 6 cm.

As we know that, areaof circle is given by:π⋅ r2where r is the radius of the circle.

Let A =π⋅ r2

As we know that, ify = f(x), then dy/dx denotes the rate of change of y with respect to x.

So, by differentiating A with respect to twe get:

\(\frac{{dA}}{{dt}} = \frac{{dA}}{{dr}} \cdot \frac{{dr}}{{dt}}\)——— (Chain Rule)

\(\Rightarrow \frac{{dA}}{{dr}} = \frac{{d\left( {π \cdot {r^2}} \right)}}{{dr}} = 2π r\)

By substituting dA/dr in dA/dt, we get

\(\Rightarrow \frac{{dA}}{{dt}} = 2π r \cdot \frac{{dr}}{{dt}}\)

By substituting r = 6 cm and dr/dt = 8 cm/sec in the above equation we get

\(\Rightarrow \frac{{dA}}{{dt}} = 2π \times 6 \times 8 = 96π \;c{m^2}/sec\)

Hence, therate of increase in the area of circular waves formedat the instant when its radius is 6 cm is 96π cm²/ sec.

Example 9: Find the rate of change of the area of a circle with respect to its radius r when r = 6 cm.

Concept;

• Area of circle is given by:π⋅ r²where r is the radius of the circle.
• If y = f(x), then dy/dx denotes the rate of change of y with respect to xits value at x = a is denoted by:\({\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}\)
• Decreasing rate is represented by negative sign whereas increasing rate is represented bypositive sign.

Calculation:

Here we have to find therate of change of the area of a circle with respect to its radius r when r = 6 cm.

As we know that, areaof circle is given by:π⋅ r2where r is the radius of the circle.

Let A =π⋅ r²

As we know that, ify = f(x), then dy/dx denotes the rate of change of y with respect to x.

So, by differentiating A with respect to r we get,

\(\Rightarrow \frac{{dA}}{{dr}} = \frac{{d\left( {π \cdot {r^2}} \right)}}{{dr}} = 2π r\)

Now we have to find the value of dA/dr at r = 6 cm i.e\({\left[ {\frac{{dA}}{{dr}}} \right]_{r\; = 6}}\)

\(\Rightarrow {\left[ {\frac{{dA}}{{dr}}} \right]_{r\; = 6}} = 2π \cdot 6 = 12π \;cm\)

Hence, therate of change of the area of a circle with respect to its radius r when r = 6 cm is 12π cm.

Example 10: If radius of circle is increasing at rate 0.5 cm/sec what is the rate of increase of its circumference?

Concept:

Rate of change of ‘x’is given by \(\rm \frac {dx}{dt}\)

Circumference of circle = 2πr

Calculation:

Here, \(\rm \frac {dr}{dt}\) = 0.5 cm/sec

Circumference of circle, C = 2πr

Now taking derivatives on both sides, we get

\(\rm \frac {dC}{dt}\) = 2π \(\rm \frac {dr}{dt}\)

= 2π (0.5)

=π cm/sec

Hence, option (4) is correct.

Example 11: Which of the following is true regarding the function f(x) = tan^-1 (cos x + sin x)?

Concept:

• \(\frac{{d\left( {{{\tan }^{ – 1}}x} \right)}}{{dx}} = \frac{1}{{1 + {x^2}}}\;\)
• Chain Rule:
• Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then\(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} \cdot \frac{{dv}}{{dx}}\)

Let f(x) be a function defined on an interval (a, b), this function is said to be a strictlyincreasing function:

• If x1< x2then f(x1) < f(x2) ∀ x1, x2∈ (a, b)
• Here,\(\frac{{dy}}{{dx}} > 0\;or\;f’\left( x \right) > 0\)

Similarly, f(x) is said to be a decreasing function:

• IfIf x1< x2thenf(x1) >f(x2) ∀ x1, x2∈ (a, b).
• Here,\(\frac{{dy}}{{dx}} < 0\;or\;f’\left( x \right) < 0\)

Calculation:

Given:f(x) = tan-1(cos x + sin x)

Let’s find out f'(x)

As we know that,\(\frac{{d\left( {{{\tan }^{ – 1}}x} \right)}}{{dx}} = \frac{1}{{1 + {x^2}}}\;\)and according to chain rule\(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} \cdot \frac{{dv}}{{dx}}\)

\(⇒ f’\left( x \right) = \frac{1}{{1 + {{\left( {\cos x + \sin x} \right)}^2}}} \cdot \frac{{d\left( {\cos x + \sin x} \right)}}{{dx}}\)

\(⇒ f’\left( x \right) = \frac{{\cos x – \sin x}}{{2 + \sin 2x}}\)

Now when 0 < x <π/4, we have cos x > sin x and sin 2x > 0

⇒ cos x – sin x > 0 and 2 + 2 sin 2x > 0

⇒ f'(x) > 0∀ x∈ (0,π/4)

As we know that for a strictly increasing function f'(x) > 0 for all x∈ (a, b)

So, the given function f(x) is astrictly increasing function on(0,π/4).

Hence option 3 is true.

Example 12: Which of the following is true regarding f(x) = x – sin x?

Concept:

Let f(x) be a function defined on an interval (a, b), this function is said to be an increasing function:

• If x1< x2then f(x1)≤ f(x2f(x1) ≤ f(x2) ∀ x1, x2∈ (a, b)
• Here,\(\frac{{dy}}{{dx}} \ge 0\;or\;f’\left( x \right) \ge 0\)

Similarly, f(x) is said to be a decreasing function:

• IfIf x1< x2thenf(x1) ≥ f(x2) ∀ x1, x2∈ (a, b).
• Here,\(\frac{{dy}}{{dx}} \le 0\;or\;f’\left( x \right) \le 0\)

Calculation:

Given:f(x) = x – sin x

Let’s calculate f'(x).

⇒ f'(x) = 1 – cos x

As we know that, – 1≤ cos x≤ 1∀ x∈ R

⇒ f'(x) = 1 – cos x≥ 0∀ x∈ R

As we know that for an increasing function say f(x) we havef'(x)≥ 0

Hence, the given function f(x) is an increasing function on R.

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