Samacheer Kalvi Class 11 Business Maths Solutions Chapter 6 Applications of Differentiation Exercise 6.4

Question 1. If \( z = (ax + b) (cy + d) \), then find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
Answer: Given the function \( z = (ax + b) (cy + d) \). To find the partial derivative with respect to \( x \), we treat \( y \) and any terms involving \( y \) as constants.
\( \frac{\partial z}{\partial x} = (cy + d) \frac{\partial}{\partial x} (ax + b) \)
\( = (cy + d) (a + 0) \)
\( = a(cy + d) \) Next, to find the partial derivative with respect to \( y \), we treat \( x \) and any terms involving \( x \) as constants.
\( \frac{\partial z}{\partial y} = (ax + b) \frac{\partial}{\partial y} (cy + d) \)
\( = (ax + b)(c + 0) \)
\( = c(ax + b) \) Partial differentiation helps us understand how a function changes when only one variable changes at a time, while others remain fixed.
In simple words: To find how \( z \) changes with \( x \), you treat \( cy + d \) as a normal number. When finding how \( z \) changes with \( y \), you treat \( ax + b \) as a normal number. You then multiply the derivative of the changing part by the constant part.

🎯 Exam Tip: When calculating partial derivatives, clearly identify which variables are treated as constants. This is key to applying the differentiation rules correctly.

Question 2. If \( u = e^{xy} \), then show that \( \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = u(x^2 + y^2) \).
Answer: Given the function \( u = e^{xy} \). First, we find the partial derivative of \( u \) with respect to \( x \), treating \( y \) as a constant:
\( \frac{\partial u}{\partial x} = y e^{xy} \) Now, we find the second partial derivative with respect to \( x \):
\( \frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial}{\partial x} (y e^{xy}) \)
\( = y \frac{\partial}{\partial x} (e^{xy}) \)
\( = y (y e^{xy}) \)
\( = y^2 e^{xy} \) ......... (1) Next, we find the partial derivative of \( u \) with respect to \( y \), treating \( x \) as a constant:
\( \frac{\partial u}{\partial y} = x e^{xy} \) Now, we find the second partial derivative with respect to \( y \):
\( \frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial}{\partial y} (x e^{xy}) \)
\( = x \frac{\partial}{\partial y} (e^{xy}) \)
\( = x (x e^{xy}) \)
\( = x^2 e^{xy} \) ......... (2) Adding (1) and (2) gives:
\( \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = y^2 e^{xy} + x^2 e^{xy} \)
\( = e^{xy}(y^2 + x^2) \) Since \( u = e^{xy} \), we can write the expression as:
\( = u(x^2 + y^2) \) Thus, it is shown that \( \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} = u(x^2 + y^2) \). Higher-order partial derivatives are important in physics and engineering, especially for analyzing wave equations or heat flow.
In simple words: Find the first and second partial derivatives of \( u \) for both \( x \) and \( y \). Add these two second derivatives together. You will see that the result matches \( u \) multiplied by \( (x^2 + y^2) \). This proves the statement.

🎯 Exam Tip: Remember to apply the chain rule correctly when differentiating partially. For a function like \( e^{xy} \), differentiating with respect to \( x \) yields \( y e^{xy} \), and with respect to \( y \) yields \( x e^{xy} \).

Question 3. Let \( u = x \cos y + y \cos x \). Verify \( \frac{\partial^{2} u}{\partial x \partial y}=\frac{\partial^{2} u}{\partial y \partial x} \).
Answer: Given the function \( u = x \cos y + y \cos x \). First, we find \( \frac{\partial^{2} u}{\partial x \partial y} \). This means we first differentiate \( u \) with respect to \( y \) and then differentiate the result with respect to \( x \). Differentiating \( u \) partially with respect to \( y \):
\( \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x \cos y) + \frac{\partial}{\partial y} (y \cos x) \)
\( = x(-\sin y) + \cos x (1) \)
\( = -x \sin y + \cos x \) Now, differentiating this result partially with respect to \( x \):
\( \frac{\partial^{2} u}{\partial x \partial y} = \frac{\partial}{\partial x} (-x \sin y + \cos x) \)
\( = -\sin y \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial x} (\cos x) \)
\( = -\sin y (1) - \sin x \)
\( = -\sin y - \sin x \) ......... (1) Next, we find \( \frac{\partial^{2} u}{\partial y \partial x} \). This means we first differentiate \( u \) with respect to \( x \) and then differentiate the result with respect to \( y \). Differentiating \( u \) partially with respect to \( x \):
\( \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x \cos y) + \frac{\partial}{\partial x} (y \cos x) \)
\( = \cos y (1) + y(-\sin x) \)
\( = \cos y - y \sin x \) Now, differentiating this result partially with respect to \( y \):
\( \frac{\partial^{2} u}{\partial y \partial x} = \frac{\partial}{\partial y} (\cos y - y \sin x) \)
\( = \frac{\partial}{\partial y} (\cos y) - \sin x \frac{\partial}{\partial y} (y) \)
\( = -\sin y - \sin x (1) \)
\( = -\sin y - \sin x \) ......... (2) From (1) and (2), we see that \( \frac{\partial^{2} u}{\partial x \partial y}=\frac{\partial^{2} u}{\partial y \partial x} \). Thus, it is verified. This equality of mixed partial derivatives is known as Clairaut's Theorem, and it holds true for most well-behaved functions.
In simple words: First, find the partial derivative of \( u \) with respect to \( y \), then differentiate that result with respect to \( x \). Next, do it the other way around: differentiate \( u \) with respect to \( x \), then differentiate that result with respect to \( y \). You should find that both final answers are the same, which means the mixed partial derivatives are equal.

🎯 Exam Tip: This verification demonstrates Clairaut's Theorem (or Schwarz's Theorem), which states that for continuous second partial derivatives, the order of differentiation does not matter. Ensure your calculations for both mixed derivatives are precise.

Question 4. Verify Euler's theorem for the function \( u = x^3 + y^3 + 3xy^2 \).
Answer: Given the function \( u = x^3 + y^3 + 3xy^2 \). To verify Euler's theorem, we first check if the function is homogeneous and determine its degree. Replace \( x \) with \( tx \) and \( y \) with \( ty \):
\( u(tx, ty) = (tx)^3 + (ty)^3 + 3(tx)(ty)^2 \)
\( = t^3x^3 + t^3y^3 + 3t x (t^2y^2) \)
\( = t^3x^3 + t^3y^3 + 3t^3xy^2 \)
\( = t^3(x^3 + y^3 + 3xy^2) \)
\( = t^3u \) Since \( u(tx, ty) = t^3u \), the function \( u \) is a homogeneous function of degree \( n = 3 \). Euler's theorem states that for a homogeneous function \( u \) of degree \( n \):
\( x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y} = nu \) In this case, it should be \( x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y} = 3u \). Now, we calculate the partial derivatives: Differentiating \( u \) partially with respect to \( x \):
\( \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x^3 + y^3 + 3xy^2) \)
\( = 3x^2 + 0 + 3y^2(1) \)
\( = 3x^2 + 3y^2 \) Multiply by \( x \):
\( x \cdot \frac{\partial u}{\partial x} = x(3x^2 + 3y^2) = 3x^3 + 3xy^2 \) ......... (1) Differentiating \( u \) partially with respect to \( y \):
\( \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x^3 + y^3 + 3xy^2) \)
\( = 0 + 3y^2 + 3x(2y) \)
\( = 3y^2 + 6xy \) Multiply by \( y \):
\( y \cdot \frac{\partial u}{\partial y} = y(3y^2 + 6xy) = 3y^3 + 6xy^2 \) ......... (2) Adding (1) and (2) gives:
\( x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y} = (3x^3 + 3xy^2) + (3y^3 + 6xy^2) \)
\( = 3x^3 + 3y^3 + 9xy^2 \) Factor out 3 from the expression:
\( = 3(x^3 + y^3 + 3xy^2) \) Since \( u = x^3 + y^3 + 3xy^2 \), we can write this as \( 3u \). Therefore, \( x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y} = 3u \). Hence, Euler's theorem is verified for the given function. Euler's theorem for homogeneous functions simplifies calculations and is often used in economics and physics to analyze functions that scale uniformly.
In simple words: First, check if the function \( u \) is "homogeneous" by replacing \( x \) with \( tx \) and \( y \) with \( ty \). If you can factor out \( t \) raised to some power (which will be 3 here), then 3 is the degree of the function. Euler's theorem says that \( x \) times the partial derivative of \( u \) with respect to \( x \), plus \( y \) times the partial derivative of \( u \) with respect to \( y \), should be equal to 3 times \( u \). Calculate both sides and show they are equal to verify the theorem.

🎯 Exam Tip: Always begin by testing for homogeneity by substituting \( tx \) and \( ty \) into the function. The power of \( t \) you can factor out will be the degree \( n \) for Euler's theorem.

Question 5. Let \( u = x^2y^3 \cos(\frac{x}{y}) \). By using Euler's theorem show that \( x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}=5 u \).
Answer: Given the function \( u = x^2y^3 \cos(\frac{x}{y}) \). To use Euler's theorem, we first determine if the function is homogeneous and find its degree. Replace \( x \) with \( tx \) and \( y \) with \( ty \):
\( u(tx, ty) = (tx)^2(ty)^3 \cos(\frac{tx}{ty}) \)
\( = t^2x^2 \cdot t^3y^3 \cos(\frac{x}{y}) \)
\( = t^{2+3} x^2y^3 \cos(\frac{x}{y}) \)
\( = t^5 x^2y^3 \cos(\frac{x}{y}) \)
\( = t^5 u \) Since \( u(tx, ty) = t^5u \), the function \( u \) is a homogeneous function of degree \( n = 5 \). According to Euler's theorem for homogeneous functions of degree \( n \):
\( x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y} = nu \) Substituting the degree \( n = 5 \) into Euler's theorem:
\( x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y} = 5u \) Thus, by using Euler's theorem, it is shown that \( x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}=5 u \). Functions that involve ratios of variables, like \( \cos(x/y) \), often lead to homogeneous functions, which simplifies their partial derivative analysis using Euler's theorem.
In simple words: To show this using Euler's theorem, first check if the function \( u \) is homogeneous. Replace \( x \) with \( tx \) and \( y \) with \( ty \). If \( t \) can be factored out with a power (which will be 5 here), then 5 is the degree of the function. Euler's theorem then directly states that \( x \) times the partial derivative of \( u \) with respect to \( x \), plus \( y \) times the partial derivative of \( u \) with respect to \( y \), will be equal to 5 times \( u \). This directly proves the statement.

🎯 Exam Tip: For functions that are products involving terms like \( \cos(\frac{x}{y}) \) or \( \log(\frac{y}{x}) \), the argument itself is of degree zero, so only the multiplied terms contribute to the overall degree of homogeneity.