Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.9

Question 1. Find \( y_2 \) for the following functions:
(i) \( y = e^{3x+2} \)
(ii) \( y = \log x + a^x \)
(iii) \( x = a \cos \theta, y = a \sin \theta \)
Answer:
(i) Given \( y = e^{3x+2} \)
To find the first derivative \( y_1 \):
\( y_1 = \frac{dy}{dx} = \frac{d}{dx} (e^{3x+2}) \)
Using the chain rule, we differentiate \( e^{u} \) as \( e^{u} \frac{du}{dx} \):
\( = e^{3x+2} \frac{d}{dx} (3x+2) \)
\( = e^{3x+2} (3(1) + 0) \)
\( = 3e^{3x+2} \)
Now, to find the second derivative \( y_2 \):
\( y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx} (3e^{3x+2}) \)
Since 3 is a constant, we take it out:
\( = 3 \frac{d}{dx} (e^{3x+2}) \)
We already found \( \frac{d}{dx} (e^{3x+2}) = 3e^{3x+2} \):
\( = 3[3e^{3x+2}] \)
\( = 9e^{3x+2} \)
Since \( y = e^{3x+2} \), we can substitute \( y \) back into the expression:
\( = 9y \)

(ii) Given \( y = \log x + a^x \)
To find the first derivative \( y_1 \):
\( y_1 = \frac{dy}{dx} = \frac{d}{dx} (\log x) + \frac{d}{dx} (a^x) \)
We know that \( \frac{d}{dx} (\log x) = \frac{1}{x} \) and \( \frac{d}{dx} (a^x) = a^x \log a \):
\( = \frac{1}{x} + a^x \log a \)
Now, to find the second derivative \( y_2 \):
\( y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx} (\frac{1}{x}) + \frac{d}{dx} (a^x \log a) \)
We know that \( \frac{d}{dx} (\frac{1}{x}) = \frac{d}{dx} (x^{-1}) = -1x^{-2} = -\frac{1}{x^2} \). The term \( \log a \) is a constant:
\( = -\frac{1}{x^2} + (\log a) \frac{d}{dx} (a^x) \)
Substitute \( \frac{d}{dx} (a^x) = a^x \log a \):
\( = -\frac{1}{x^2} + (\log a) (a^x \log a) \)
\( = -\frac{1}{x^2} + a^x (\log a)^2 \)

(iii) Given \( x = a \cos \theta \) and \( y = a \sin \theta \)
We need to find \( \frac{d^2y}{dx^2} \). First, find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).
\( \frac{dx}{d\theta} = \frac{d}{d\theta} (a \cos \theta) = a(-\sin \theta) = -a \sin \theta \) ..... (i)
\( \frac{dy}{d\theta} = \frac{d}{d\theta} (a \sin \theta) = a(\cos \theta) \)
Now, find the first derivative \( y_1 = \frac{dy}{dx} \) using the chain rule: \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).
\( y_1 = \frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta} \)
\( = -\cot \theta \)
To find the second derivative \( y_2 = \frac{d^2y}{dx^2} \), we differentiate \( y_1 \) with respect to \( x \). Since \( y_1 \) is a function of \( \theta \), we use the chain rule again: \( \frac{d}{dx} = \frac{d}{d\theta} \cdot \frac{d\theta}{dx} \).
\( y_2 = \frac{d^2y}{dx^2} = \frac{d}{d\theta} (-\cot \theta) \cdot \frac{d\theta}{dx} \)
We know that \( \frac{d}{d\theta} (-\cot \theta) = -(-\text{cosec}^2 \theta) = \text{cosec}^2 \theta \).
From equation (i), \( \frac{dx}{d\theta} = -a \sin \theta \), so \( \frac{d\theta}{dx} = \frac{1}{-a \sin \theta} \).
\( = (\text{cosec}^2 \theta) \cdot \frac{1}{-a \sin \theta} \)
We know that \( \text{cosec} \theta = \frac{1}{\sin \theta} \), so \( \frac{1}{\sin \theta} = \text{cosec} \theta \).
\( = \frac{\text{cosec}^2 \theta}{-a \sin \theta} \)
\( = \frac{\text{cosec}^2 \theta}{-a} \cdot \text{cosec} \theta \)
\( = -\frac{1}{a} \text{cosec}^3 \theta \)
In simple words: For each function, we first find the rate of change once (called the first derivative or \( y_1 \)). Then, we find the rate of change of that first result (called the second derivative or \( y_2 \)). This shows how the change itself is changing. For part (iii), we used a special method for functions that depend on a third variable, like \( \theta \).

🎯 Exam Tip: Remember the basic differentiation formulas for \( e^x, \log x, a^x, \sin x, \cos x \), and always apply the chain rule correctly when differentiating composite functions.

Question 2. If \( y = 500e^{7x} + 600e^{-7x} \), then show that \( y_2 – 49y = 0 \).
Answer: Given the function:
\( y = 500e^{7x} + 600e^{-7x} \)
First, find the first derivative \( y_1 \):
\( y_1 = \frac{dy}{dx} = \frac{d}{dx} (500e^{7x}) + \frac{d}{dx} (600e^{-7x}) \)
Using the chain rule, \( \frac{d}{dx} (e^{ax}) = ae^{ax} \):
\( = 500(7e^{7x}) + 600(-7e^{-7x}) \)
\( = 3500e^{7x} - 4200e^{-7x} \)
Next, find the second derivative \( y_2 \):
\( y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx} (3500e^{7x}) - \frac{d}{dx} (4200e^{-7x}) \)
Applying the chain rule again:
\( = 3500(7e^{7x}) - 4200(-7e^{-7x}) \)
\( = 24500e^{7x} + 29400e^{-7x} \)
We can factor out 49 from both terms:
\( = 49(500e^{7x}) + 49(600e^{-7x}) \)
Factor out 49 from the entire expression:
\( = 49[500e^{7x} + 600e^{-7x}] \)
Notice that the expression inside the brackets is the original function \( y \):
\( y_2 = 49y \)
Rearrange the equation to show the desired result:
\( y_2 - 49y = 0 \)
In simple words: We find the first derivative and then the second derivative of the given equation. After some calculations, we see that the second derivative is exactly 49 times the original function. When we move the \( 49y \) to the other side, we get zero, proving the statement.

🎯 Exam Tip: When proving an identity involving derivatives, it's often easiest to calculate the required derivatives separately (like \( y_1 \) and \( y_2 \)) and then substitute them into the given equation to check if it holds true.

Question 3. If \( y = 2 + \log x \), then show that \( xy_2 + y_1 = 0 \).
Answer: Given the function:
\( y = 2 + \log x \)
First, find the first derivative \( y_1 \):
\( y_1 = \frac{dy}{dx} = \frac{d}{dx} (2) + \frac{d}{dx} (\log x) \)
The derivative of a constant (2) is 0, and the derivative of \( \log x \) is \( \frac{1}{x} \):
\( y_1 = 0 + \frac{1}{x} \)
\( y_1 = \frac{1}{x} \)
Next, find the second derivative \( y_2 \):
\( y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx} (\frac{1}{x}) \)
We know that \( \frac{d}{dx} (x^{-1}) = -1x^{-2} = -\frac{1}{x^2} \):
\( y_2 = -\frac{1}{x^2} \)
Now, substitute \( y_1 \) and \( y_2 \) into the expression we need to show: \( xy_2 + y_1 \).
\( xy_2 + y_1 = x(-\frac{1}{x^2}) + \frac{1}{x} \)
\( = -\frac{x}{x^2} + \frac{1}{x} \)
\( = -\frac{1}{x} + \frac{1}{x} \)
\( = 0 \)
Thus, we have shown that \( xy_2 + y_1 = 0 \).
In simple words: We found the first and second rates of change for the given function. Then we put these results into the equation \( xy_2 + y_1 \). When we added them up, the total was zero, which proves the statement.

🎯 Exam Tip: Be careful with the signs and powers when differentiating terms like \( \frac{1}{x} \). Always simplify expressions completely after substitution to reach the desired result.

Question 4. If \( y = a \cos mx + b \sin mx \), then show that \( y_2 + m^2y = 0 \).
Answer: Given the function:
\( y = a \cos mx + b \sin mx \)
First, find the first derivative \( y_1 \):
\( y_1 = \frac{dy}{dx} = \frac{d}{dx} (a \cos mx) + \frac{d}{dx} (b \sin mx) \)
Using the chain rule, \( \frac{d}{dx} (\cos u) = -\sin u \frac{du}{dx} \) and \( \frac{d}{dx} (\sin u) = \cos u \frac{du}{dx} \):
\( = a(-\sin mx) \cdot m + b(\cos mx) \cdot m \)
\( = -am \sin mx + bm \cos mx \)
Next, find the second derivative \( y_2 \):
\( y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx} (-am \sin mx) + \frac{d}{dx} (bm \cos mx) \)
Applying the chain rule again:
\( = -am(\cos mx) \cdot m + bm(-\sin mx) \cdot m \)
\( = -am^2 \cos mx - bm^2 \sin mx \)
Factor out \( -m^2 \) from both terms:
\( = -m^2 [a \cos mx + b \sin mx] \)
Notice that the expression inside the brackets is the original function \( y \):
\( y_2 = -m^2y \)
Rearrange the equation to show the desired result:
\( y_2 + m^2y = 0 \)
In simple words: We calculated the first and second derivatives of the given trigonometric function. After finding the second derivative, we saw that it was equal to \( -m^2 \) times the original function. Moving this term to the other side gives us the required equation, showing the relationship between the function and its second derivative.

🎯 Exam Tip: Always pay close attention to the constant \( m \) and its signs when applying the chain rule to trigonometric functions like \( \cos mx \) and \( \sin mx \).

Question 5. If \( y = \left(x+\sqrt{1+x^{2}}\right)^{m} \), then show that \( (1 + x^2) y_2 + xy_1 – m^2y = 0 \).
Answer: Given the function:
\( y = (x + \sqrt{1+x^2})^m \)
To simplify differentiation, we can use implicit differentiation or logarithmic differentiation, but the provided steps follow a direct method.
First, find the first derivative \( y_1 \):
\( y_1 = \frac{dy}{dx} = m(x + \sqrt{1+x^2})^{m-1} \cdot \frac{d}{dx} (x + \sqrt{1+x^2}) \)
Differentiate the term inside the bracket: \( \frac{d}{dx} (x) = 1 \) and \( \frac{d}{dx} (\sqrt{1+x^2}) = \frac{1}{2\sqrt{1+x^2}} \cdot \frac{d}{dx} (1+x^2) = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}} \).
So, \( \frac{d}{dx} (x + \sqrt{1+x^2}) = 1 + \frac{x}{\sqrt{1+x^2}} \)
\( = 1 + \frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}} \)
Substitute this back into the expression for \( y_1 \):
\( y_1 = m(x + \sqrt{1+x^2})^{m-1} \cdot \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}} \)
We can rewrite \( (x + \sqrt{1+x^2})^{m-1} \cdot (x + \sqrt{1+x^2})^1 \) as \( (x + \sqrt{1+x^2})^m = y \).
So, \( y_1 = \frac{m(x + \sqrt{1+x^2})^m}{\sqrt{1+x^2}} \)
\( y_1 = \frac{my}{\sqrt{1+x^2}} \)
To remove the square root, square both sides:
\( y_1^2 = \frac{m^2y^2}{1+x^2} \)
Rearrange the terms:
\( (1+x^2)y_1^2 = m^2y^2 \)
Now, differentiate both sides with respect to \( x \) again. Use the product rule for the left side and chain rule for \( y^2 \):
\( \frac{d}{dx} [(1+x^2)y_1^2] = \frac{d}{dx} [m^2y^2] \)
\( (1+x^2) \cdot 2y_1 \frac{dy_1}{dx} + y_1^2 \cdot \frac{d}{dx}(1+x^2) = m^2 \cdot 2y \frac{dy}{dx} \)
\( (1+x^2) \cdot 2y_1 y_2 + y_1^2 (2x) = m^2 \cdot 2yy_1 \)
Divide the entire equation by \( 2y_1 \) (assuming \( y_1 \neq 0 \)):
\( (1+x^2)y_2 + xy_1 = m^2y \)
Rearrange the terms to get the desired form:
\( (1+x^2)y_2 + xy_1 - m^2y = 0 \)
In simple words: We first found the first rate of change (\( y_1 \)) for the complex function. Then, to get rid of the square root, we squared both sides and rearranged the equation. After that, we found the second rate of change (\( y_2 \)) by differentiating again. Finally, by putting all the parts together and simplifying, we proved the given equation.

🎯 Exam Tip: For complex functions, sometimes it's easier to manipulate the first derivative (e.g., squaring to remove square roots) before taking the second derivative to simplify the process and avoid errors.

Question 6. If \( y = \sin(\log x) \), then show that \( x^2y_2 + xy_1 + y = 0 \).
Answer: Given the function:
\( y = \sin(\log x) \)
First, find the first derivative \( y_1 \):
\( y_1 = \frac{dy}{dx} = \frac{d}{dx} (\sin(\log x)) \)
Using the chain rule, \( \frac{d}{dx} (\sin u) = \cos u \frac{du}{dx} \):
\( y_1 = \cos(\log x) \cdot \frac{d}{dx} (\log x) \)
\( y_1 = \cos(\log x) \cdot \frac{1}{x} \)
Multiply both sides by \( x \) to simplify:
\( xy_1 = \cos(\log x) \)
Next, differentiate both sides with respect to \( x \) again. Use the product rule for the left side:
\( \frac{d}{dx} (xy_1) = \frac{d}{dx} (\cos(\log x)) \)
\( x \frac{dy_1}{dx} + y_1 \frac{d}{dx} (x) = -\sin(\log x) \cdot \frac{d}{dx} (\log x) \)
\( xy_2 + y_1(1) = -\sin(\log x) \cdot \frac{1}{x} \)
\( xy_2 + y_1 = -\frac{\sin(\log x)}{x} \)
Multiply the entire equation by \( x \) to remove the fraction:
\( x(xy_2 + y_1) = -\sin(\log x) \)
\( x^2y_2 + xy_1 = -\sin(\log x) \)
Recall that \( y = \sin(\log x) \), so substitute \( -y \) for \( -\sin(\log x) \):
\( x^2y_2 + xy_1 = -y \)
Rearrange the terms to get the desired form:
\( x^2y_2 + xy_1 + y = 0 \)
In simple words: We took the first and second derivatives of the function, being careful with the chain rule. After finding these derivatives, we substituted them back into the given equation. We then simplified the terms, and the equation became zero, which proved the statement.

🎯 Exam Tip: When faced with an equation involving derivatives and a function, it's often helpful to multiply by \( x \) or \( x^2 \) at intermediate steps to clear denominators and simplify further differentiation using the product rule.