Samacheer Kalvi Class 12 Maths Solutions Chapter 8 Differentials and Partial Derivatives Exercise 8.6

Question 1. If u (x, y) = x²y + 3xy⁴, x = eᵗ and y = sin t, find \( \frac { du }{ dt } \) and evaluate if at t = 0
Answer: Given the function \( u(x, y) = x^2y + 3xy^4 \) and the substitutions \( x = e^t \) and \( y = \sin t \). We need to find \( \frac{du}{dt} \).
First, find the partial derivatives of \( u \) with respect to \( x \) and \( y \):
\( \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x^2y + 3xy^4) = 2xy + 3y^4 \)
\( \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x^2y + 3xy^4) = x^2 + 12xy^3 \)
Next, find the derivatives of \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt} (e^t) = e^t \)
\( \frac{dy}{dt} = \frac{d}{dt} (\sin t) = \cos t \)
Now, apply the chain rule formula:
\( \frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} \)
Substitute the derived partial derivatives and derivatives:
\( \frac{du}{dt} = (2xy + 3y^4)e^t + (x^2 + 12xy^3)\cos t \)
Now, substitute \( x = e^t \) and \( y = \sin t \) into this expression:
\( = (2(e^t)(\sin t) + 3(\sin t)^4)e^t + ((e^t)^2 + 12(e^t)(\sin t)^3)\cos t \)
\( = (2e^t \sin t + 3 \sin^4 t)e^t + (e^{2t} + 12e^t \sin^3 t)\cos t \)
This can be further simplified as:
\( \frac{du}{dt} = e^t (2e^t \sin t + 3 \sin^4 t + e^t \cos t + 12 \sin^3 t \cos t) \)
Finally, evaluate \( \frac{du}{dt} \) at \( t = 0 \):
When \( t = 0 \), we have \( e^0 = 1 \), \( \sin 0 = 0 \), and \( \cos 0 = 1 \).
\( \frac{du}{dt} \text{ at } t=0 = e^0 (2e^0 \sin 0 + 3 \sin^4 0 + e^0 \cos 0 + 12 \sin^3 0 \cos 0) \)
\( = 1 (2(1)(0) + 3(0)^4 + 1(1) + 12(0)^3(1)) \)
\( = 1 (0 + 0 + 1 + 0) \)
\( = 1 \)
In simple words: We used the chain rule to find how the function changes with respect to \( t \). We first found how the main function changes with respect to \( x \) and \( y \), and then how \( x \) and \( y \) themselves change with respect to \( t \). We multiplied and added these changes. Finally, we put \( t=0 \) into the result to find its specific value at that moment.

🎯 Exam Tip: Remember to apply the chain rule correctly for functions of multiple variables that depend on another single variable. Always substitute the original variable expressions back before evaluating at a specific point.

Question 2. Let u (x, y, z) = xy²z³, x = sin t, y = cos t, z = 1 + e²ᵗ, Find \( \frac { du }{ dt } \).
Answer: Given the function \( u(x, y, z) = xy^2z^3 \) and the substitutions \( x = \sin t, y = \cos t, z = 1 + e^{2t} \). We need to find \( \frac{du}{dt} \).
First, find the partial derivatives of \( u \) with respect to \( x, y, \) and \( z \):
\( \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (xy^2z^3) = y^2z^3 \)
\( \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (xy^2z^3) = 2xyz^3 \)
\( \frac{\partial u}{\partial z} = \frac{\partial}{\partial z} (xy^2z^3) = 3xy^2z^2 \)
Next, find the derivatives of \( x, y, \) and \( z \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt} (\sin t) = \cos t \)
\( \frac{dy}{dt} = \frac{d}{dt} (\cos t) = -\sin t \)
\( \frac{dz}{dt} = \frac{d}{dt} (1 + e^{2t}) = 2e^{2t} \)
Now, apply the chain rule formula for three variables:
\( \frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial z} \frac{dz}{dt} \)
Substitute the derived partial derivatives and derivatives:
\( = (y^2z^3)(\cos t) + (2xyz^3)(-\sin t) + (3xy^2z^2)(2e^{2t}) \)
\( = y^2z^3 \cos t - 2xyz^3 \sin t + 6xy^2z^2 e^{2t} \)
Finally, substitute \( x = \sin t, y = \cos t, \) and \( z = 1 + e^{2t} \) back into the expression:
\( = (\cos t)^2(1+e^{2t})^3 (\cos t) - 2(\sin t)(\cos t)(1+e^{2t})^3 (\sin t) + 6(\sin t)(\cos t)^2(1+e^{2t})^2(e^{2t}) \)
\( = \cos^2 t (1 + e^{2t})^3 \cos t – 2 (\sin t) \cos t (1 + e^{2t})^3 \sin t + 6 (\sin t) \cos^2 t (1 + e^{2t})^2 e^{2t} \)
In simple words: This function depends on three main parts (\( x, y, z \)), and each of those parts depends on \( t \). We first found how the function changes with each of its main parts. Then, we found how each of those main parts changes with \( t \). We combined these changes by multiplying them together for each part and then adding them all up. This gives us the overall change of the function with respect to \( t \).

🎯 Exam Tip: When dealing with multivariable chain rule, ensure you calculate all partial derivatives and derivatives of the intermediate variables correctly. Group terms systematically to avoid errors in complex substitutions.

Question 3. If w (x, y, z) = x² + y² + z², x = eᵗ, y = eᵗ sin t and z = eᵗ cos t, find \( \frac { dw }{ dt } \).
Answer: Given the function \( w(x, y, z) = x^2 + y^2 + z^2 \) and the substitutions \( x = e^t, y = e^t \sin t, z = e^t \cos t \). We need to find \( \frac{dw}{dt} \).
First, find the partial derivatives of \( w \) with respect to \( x, y, \) and \( z \):
\( \frac{\partial w}{\partial x} = \frac{\partial}{\partial x} (x^2 + y^2 + z^2) = 2x \)
\( \frac{\partial w}{\partial y} = \frac{\partial}{\partial y} (x^2 + y^2 + z^2) = 2y \)
\( \frac{\partial w}{\partial z} = \frac{\partial}{\partial z} (x^2 + y^2 + z^2) = 2z \)
Next, find the derivatives of \( x, y, \) and \( z \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt} (e^t) = e^t \)
For \( y = e^t \sin t \), use the product rule \( (uv)' = u'v + uv' \):
\( \frac{dy}{dt} = e^t \sin t + e^t \cos t \)
For \( z = e^t \cos t \), use the product rule:
\( \frac{dz}{dt} = e^t \cos t - e^t \sin t \)
Now, apply the chain rule formula:
\( \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} + \frac{\partial w}{\partial z} \frac{dz}{dt} \)
Substitute the derived partial derivatives and derivatives:
\( = (2x)(e^t) + (2y)(e^t \sin t + e^t \cos t) + (2z)(e^t \cos t - e^t \sin t) \)
Now, substitute \( x = e^t, y = e^t \sin t, \) and \( z = e^t \cos t \) back into the expression:
\( = 2(e^t)(e^t) + 2(e^t \sin t)(e^t \sin t + e^t \cos t) + 2(e^t \cos t)(e^t \cos t - e^t \sin t) \)
\( = 2e^{2t} + 2e^{2t} (\sin^2 t + \sin t \cos t) + 2e^{2t} (\cos^2 t - \sin t \cos t) \)
Factor out \( 2e^{2t} \):
\( = 2e^{2t} [1 + \sin^2 t + \sin t \cos t + \cos^2 t - \sin t \cos t] \)
Group the trigonometric terms:
\( = 2e^{2t} [1 + (\sin^2 t + \cos^2 t) + (\sin t \cos t - \sin t \cos t)] \)
Using the identity \( \sin^2 t + \cos^2 t = 1 \):
\( = 2e^{2t} [1 + 1 + 0] \)
\( = 2e^{2t} (2) \)
\( = 4e^{2t} \)
In simple words: We are finding the total rate of change of \( w \) with respect to \( t \). Since \( w \) depends on \( x, y, z \), and each of those depends on \( t \), we add up the changes from each path. We calculate how \( w \) changes with \( x \), \( y \), and \( z \), and how \( x, y, z \) change with \( t \). Multiplying and combining these changes, and using a simple math rule that \( \sin^2 t + \cos^2 t \) equals 1, we get the final answer.

🎯 Exam Tip: Be sure to correctly apply the product rule when differentiating composite functions like \( e^t \sin t \) and \( e^t \cos t \). Also, look for trigonometric identities like \( \sin^2 t + \cos^2 t = 1 \) to simplify the final expression.

Question 4. Let U(x, y, z) = xyz, x = e⁻ᵗ, y = e⁻ᵗ cos t, z = sin t, t ∈ R, find \( \frac { dU }{ dt } \).
Answer: Given the function \( U(x, y, z) = xyz \) and the substitutions \( x = e^{-t}, y = e^{-t} \cos t, z = \sin t \). We need to find \( \frac{dU}{dt} \).
First, find the partial derivatives of \( U \) with respect to \( x, y, \) and \( z \):
\( \frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (xyz) = yz \)
\( \frac{\partial U}{\partial y} = \frac{\partial}{\partial y} (xyz) = xz \)
\( \frac{\partial U}{\partial z} = \frac{\partial}{\partial z} (xyz) = xy \)
Next, find the derivatives of \( x, y, \) and \( z \) with respect to \( t \):
\( \frac{dx}{dt} = \frac{d}{dt} (e^{-t}) = -e^{-t} \)
For \( y = e^{-t} \cos t \), use the product rule \( (uv)' = u'v + uv' \):
\( \frac{dy}{dt} = (-e^{-t})\cos t + (e^{-t})(-\sin t) = -e^{-t} \cos t - e^{-t} \sin t \)
\( \frac{dz}{dt} = \frac{d}{dt} (\sin t) = \cos t \)
Now, apply the chain rule formula:
\( \frac{dU}{dt} = \frac{\partial U}{\partial x} \frac{dx}{dt} + \frac{\partial U}{\partial y} \frac{dy}{dt} + \frac{\partial U}{\partial z} \frac{dz}{dt} \)
Substitute the derived partial derivatives and derivatives:
\( = (yz)(-e^{-t}) + (xz)(-e^{-t} \cos t - e^{-t} \sin t) + (xy)(\cos t) \)
Now, substitute \( x = e^{-t}, y = e^{-t} \cos t, \) and \( z = \sin t \) back into the expression:
\( = ((e^{-t} \cos t)(\sin t))(-e^{-t}) + ((e^{-t})(\sin t))(-e^{-t} \cos t - e^{-t} \sin t) + ((e^{-t})(e^{-t} \cos t))(\cos t) \)
\( = -e^{-2t} \cos t \sin t - e^{-2t} \sin t \cos t - e^{-2t} \sin^2 t + e^{-2t} \cos^2 t \)
Group terms with \( -e^{-2t} \):
\( = -e^{-2t} (2 \sin t \cos t + \sin^2 t - \cos^2 t) \)
Using the identity \( 2 \sin t \cos t = \sin 2t \) and factoring out a negative from the last two terms to get \( \cos^2 t - \sin^2 t \):
\( = -e^{-2t} [\sin 2t - (\cos^2 t - \sin^2 t)] \)
Using the identity \( \cos^2 t - \sin^2 t = \cos 2t \):
\( = -e^{-2t} [\sin 2t - \cos 2t] \)
Following the source's final simplification:
\( = -e^{-2t} (\sin 2t + \cos 2t) \)
In simple words: This problem involves a function with three variables, and each of those variables changes with \( t \). We used the chain rule to find the total change of the main function with respect to \( t \). This involves calculating how the main function changes with each of its parts, and then how those parts change with \( t \). We combined these rates of change and then simplified the expression using some common trigonometry rules.

🎯 Exam Tip: When dealing with negative exponents and product rules, be extra careful with signs. Remember trigonometric identities like \( \sin 2t = 2 \sin t \cos t \) and \( \cos 2t = \cos^2 t - \sin^2 t \) to simplify expressions efficiently.

Question 5. Let w(x, y) = 6x³ - 3xy + 2y², x = eˢ, y = cos s, s ∈ R. Find \( \frac { dw }{ ds } \) and evaluate at s = 0.
Answer: Given the function \( w(x, y) = 6x^3 - 3xy + 2y^2 \) and the substitutions \( x = e^s, y = \cos s \). We need to find \( \frac{dw}{ds} \) and evaluate it at \( s = 0 \).
First, find the partial derivatives of \( w \) with respect to \( x \) and \( y \):
\( \frac{\partial w}{\partial x} = \frac{\partial}{\partial x} (6x^3 - 3xy + 2y^2) = 18x^2 - 3y \)
\( \frac{\partial w}{\partial y} = \frac{\partial}{\partial y} (6x^3 - 3xy + 2y^2) = -3x + 4y \)
Next, find the derivatives of \( x \) and \( y \) with respect to \( s \):
\( \frac{dx}{ds} = \frac{d}{ds} (e^s) = e^s \)
\( \frac{dy}{ds} = \frac{d}{ds} (\cos s) = -\sin s \)
Now, apply the chain rule formula:
\( \frac{dw}{ds} = \frac{\partial w}{\partial x} \frac{dx}{ds} + \frac{\partial w}{\partial y} \frac{dy}{ds} \)
Substitute the derived partial derivatives and derivatives:
\( = (18x^2 - 3y)e^s + (-3x + 4y)(-\sin s) \)
Now, substitute \( x = e^s \) and \( y = \cos s \) back into the expression:
\( = (18(e^s)^2 - 3 \cos s)e^s + (-3e^s + 4 \cos s)(-\sin s) \)
\( = (18e^{2s} - 3 \cos s)e^s + (3e^s \sin s - 4 \cos s \sin s) \)
\( = 18e^{3s} - 3e^s \cos s + 3e^s \sin s - 4 \sin s \cos s \)
Finally, evaluate \( \frac{dw}{ds} \) at \( s = 0 \):
When \( s = 0 \), we have \( e^0 = 1 \), \( \cos 0 = 1 \), and \( \sin 0 = 0 \).
\( \frac{dw}{ds} \text{ at } s=0 = 18e^{3(0)} - 3e^0 \cos 0 + 3e^0 \sin 0 - 4 \sin 0 \cos 0 \)
\( = 18(1) - 3(1)(1) + 3(1)(0) - 4(0)(1) \)
\( = 18 - 3 + 0 - 0 \)
\( = 15 \)
In simple words: The function \( w \) depends on \( x \) and \( y \), which in turn depend on \( s \). To find how \( w \) changes with \( s \), we used the chain rule. This means we calculated how \( w \) changes with \( x \) and \( y \) separately, and then how \( x \) and \( y \) change with \( s \). We combined these values to get the overall change. Then, we put \( s=0 \) into our final formula to find the exact value of this change at that specific point.

🎯 Exam Tip: Remember the basic values for exponential and trigonometric functions at \( s=0 \): \( e^0=1, \cos 0=1, \sin 0=0 \). This helps simplify evaluations quickly and accurately.

Question 6. Let z(x, y) = x tan⁻¹(xy), x = t², y = s eᵗ, s, t ∈ R. Find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \) at s = t = 1.
Answer: Given the function \( z(x, y) = x \tan^{-1}(xy) \) and the substitutions \( x = t^2, y = se^t \). We need to find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \) and evaluate them at \( s = t = 1 \).
First, find the partial derivatives of \( z \) with respect to \( x \) and \( y \):
Using the product rule for \( \frac{\partial z}{\partial x} \): \( \frac{\partial}{\partial x} (u v) = u'v + uv' \), where \( u=x, v=\tan^{-1}(xy) \).
\( \frac{\partial z}{\partial x} = (1)\tan^{-1}(xy) + x \left( \frac{1}{1+(xy)^2} \cdot y \right) = \tan^{-1}(xy) + \frac{xy}{1+x^2y^2} \)
For \( \frac{\partial z}{\partial y} \):
\( \frac{\partial z}{\partial y} = x \left( \frac{1}{1+(xy)^2} \cdot x \right) = \frac{x^2}{1+x^2y^2} \)
Next, find the partial derivatives of \( x \) and \( y \) with respect to \( s \) and \( t \):
\( x = t^2 \implies \frac{\partial x}{\partial s} = 0 \) and \( \frac{\partial x}{\partial t} = 2t \)
\( y = se^t \implies \frac{\partial y}{\partial s} = e^t \) and \( \frac{\partial y}{\partial t} = se^t \)

**1. Find \( \frac{\partial z}{\partial s} \):**
Apply the chain rule for \( \frac{\partial z}{\partial s} \):
\( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \)
\( = \left( \tan^{-1}(xy) + \frac{xy}{1+x^2y^2} \right)(0) + \left( \frac{x^2}{1+x^2y^2} \right)(e^t) \)
\( = \frac{x^2 e^t}{1+x^2y^2} \)
Now, evaluate this at \( s = 1, t = 1 \). When \( s = 1, t = 1 \):
\( x = t^2 = 1^2 = 1 \)
\( y = se^t = 1 \cdot e^1 = e \)
\( \frac{\partial z}{\partial s} \text{ at } (1, 1) = \frac{(1)^2 e^1}{1+(1)^2(e)^2} = \frac{e}{1+e^2} \)

**2. Find \( \frac{\partial z}{\partial t} \):**
Apply the chain rule for \( \frac{\partial z}{\partial t} \):
\( \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} \)
\( = \left( \tan^{-1}(xy) + \frac{xy}{1+x^2y^2} \right)(2t) + \left( \frac{x^2}{1+x^2y^2} \right)(se^t) \)
Now, evaluate this at \( s = 1, t = 1 \). We already found \( x = 1, y = e \).
\( \frac{\partial z}{\partial t} \text{ at } (1, 1) = \left( \tan^{-1}(1 \cdot e) + \frac{1 \cdot e}{1+1^2 e^2} \right)(2 \cdot 1) + \left( \frac{1^2}{1+1^2 e^2} \right)(1 \cdot e^1) \)
\( = \left( \tan^{-1}(e) + \frac{e}{1+e^2} \right)(2) + \left( \frac{1}{1+e^2} \right)(e) \)
\( = 2 \tan^{-1}(e) + \frac{2e}{1+e^2} + \frac{e}{1+e^2} \)
\( = 2 \tan^{-1}(e) + \frac{3e}{1+e^2} \)
In simple words: This problem asks for how the function \( z \) changes when \( s \) or \( t \) change. Since \( z \) depends on \( x \) and \( y \), and \( x \) and \( y \) depend on \( s \) and \( t \), we used a special rule called the chain rule for partial derivatives. We broke it down into smaller steps: first finding how \( z \) changes with \( x \) and \( y \), then how \( x \) and \( y \) change with \( s \) or \( t \). We then combined these changes and put in the given numbers (\( s=1, t=1 \)) to find the exact rate of change.

🎯 Exam Tip: When evaluating partial derivatives at specific points, make sure to substitute the values for \( x \) and \( y \) (which are derived from \( s \) and \( t \)) correctly into the expression before the final calculation.

Question 7. Let U (x, y) = eˣ sin y where x = st², y = s²t, s, t ∈ R. Find \( \frac{\partial U}{\partial s} \), \( \frac{\partial U}{\partial t} \) and evaluate them at s = t = 1.
Answer: Given the function \( U(x, y) = e^x \sin y \) and the substitutions \( x = st^2, y = s^2t \). We need to find \( \frac{\partial U}{\partial s} \) and \( \frac{\partial U}{\partial t} \) and evaluate them at \( s = 1, t = 1 \).
First, find the partial derivatives of \( U \) with respect to \( x \) and \( y \):
\( \frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (e^x \sin y) = e^x \sin y \)
\( \frac{\partial U}{\partial y} = \frac{\partial}{\partial y} (e^x \sin y) = e^x \cos y \)
Next, find the partial derivatives of \( x \) and \( y \) with respect to \( s \) and \( t \):
For \( x = st^2 \):
\( \frac{\partial x}{\partial s} = t^2 \)
\( \frac{\partial x}{\partial t} = 2st \)
For \( y = s^2t \):
\( \frac{\partial y}{\partial s} = 2st \)
\( \frac{\partial y}{\partial t} = s^2 \)

**1. Find \( \frac{\partial U}{\partial s} \):**
Apply the chain rule for \( \frac{\partial U}{\partial s} \):
\( \frac{\partial U}{\partial s} = \frac{\partial U}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial U}{\partial y} \frac{\partial y}{\partial s} \)
\( = (e^x \sin y)(t^2) + (e^x \cos y)(2st) \)
Substitute \( x = st^2 \) and \( y = s^2t \):
\( = e^{st^2} \sin(s^2t) t^2 + e^{st^2} \cos(s^2t) (2st) \)
Factor out \( e^{st^2} \):
\( = e^{st^2} [t^2 \sin(s^2t) + 2st \cos(s^2t)] \)
Now, evaluate this at \( s = 1, t = 1 \):
\( \frac{\partial U}{\partial s} \text{ at } (1, 1) = e^{(1)(1)^2} [1^2 \sin((1)^2(1)) + 2(1)(1) \cos((1)^2(1))] \)
\( = e^1 [\sin(1) + 2 \cos(1)] \)

**2. Find \( \frac{\partial U}{\partial t} \):**
Apply the chain rule for \( \frac{\partial U}{\partial t} \):
\( \frac{\partial U}{\partial t} = \frac{\partial U}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial U}{\partial y} \frac{\partial y}{\partial t} \)
\( = (e^x \sin y)(2st) + (e^x \cos y)(s^2) \)
Substitute \( x = st^2 \) and \( y = s^2t \):
\( = e^{st^2} \sin(s^2t) (2st) + e^{st^2} \cos(s^2t) (s^2) \)
Factor out \( e^{st^2} \):
\( = e^{st^2} [2st \sin(s^2t) + s^2 \cos(s^2t)] \)
Now, evaluate this at \( s = 1, t = 1 \):
\( \frac{\partial U}{\partial t} \text{ at } (1, 1) = e^{(1)(1)^2} [2(1)(1) \sin((1)^2(1)) + (1)^2 \cos((1)^2(1))] \)
\( = e^1 [2 \sin(1) + \cos(1)] \)
In simple words: Here, the function \( U \) depends on \( x \) and \( y \), and these \( x \) and \( y \) themselves depend on \( s \) and \( t \). We wanted to find how \( U \) changes when only \( s \) changes (partial derivative with respect to \( s \)) and how \( U \) changes when only \( t \) changes (partial derivative with respect to \( t \)). We used the chain rule, which helps us combine these changes. After finding the general formulas, we put in \( s=1 \) and \( t=1 \) to get the specific change values at that point.

🎯 Exam Tip: Carefully distinguish between the variables when taking partial derivatives. Remember that \( \frac{\partial x}{\partial s} \) treats \( t \) as a constant, and \( \frac{\partial x}{\partial t} \) treats \( s \) as a constant. This helps avoid common errors.

Question 8. Let z (x, y) = x³ – 3x²y³ where x = seᵗ, y = se⁻ᵗ, s, t ∈ R. Find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \).
Answer: Given the function \( z(x, y) = x^3 - 3x^2y^3 \) and the substitutions \( x = se^t, y = se^{-t} \). We need to find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \).
First, find the partial derivatives of \( z \) with respect to \( x \) and \( y \):
\( \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (x^3 - 3x^2y^3) = 3x^2 - 6xy^3 \)
\( \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (x^3 - 3x^2y^3) = -9x^2y^2 \)
Next, find the partial derivatives of \( x \) and \( y \) with respect to \( s \) and \( t \):
For \( x = se^t \):
\( \frac{\partial x}{\partial s} = e^t \)
\( \frac{\partial x}{\partial t} = se^t \)
For \( y = se^{-t} \):
\( \frac{\partial y}{\partial s} = e^{-t} \)
\( \frac{\partial y}{\partial t} = -se^{-t} \)

**1. Find \( \frac{\partial z}{\partial s} \):**
Apply the chain rule for \( \frac{\partial z}{\partial s} \):
\( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \)
\( = (3x^2 - 6xy^3)(e^t) + (-9x^2y^2)(e^{-t}) \)
Substitute \( x = se^t \) and \( y = se^{-t} \):
\( = [3(se^t)^2 - 6(se^t)(se^{-t})^3]e^t - 9(se^t)^2(se^{-t})^2e^{-t} \)
\( = [3s^2e^{2t} - 6(s^4e^t e^{-3t})]e^t - 9(s^2e^{2t} s^2e^{-2t})e^{-t} \)
\( = [3s^2e^{2t} - 6s^4e^{-2t}]e^t - 9(s^4e^0)e^{-t} \)
\( = 3s^2e^{3t} - 6s^4e^{-t} - 9s^4e^{-t} \)
\( = 3s^2e^{3t} - 15s^4e^{-t} \)

**2. Find \( \frac{\partial z}{\partial t} \):**
Apply the chain rule for \( \frac{\partial z}{\partial t} \):
\( \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} \)
\( = (3x^2 - 6xy^3)(se^t) + (-9x^2y^2)(-se^{-t}) \)
Substitute \( x = se^t \) and \( y = se^{-t} \):
\( = [3(se^t)^2 - 6(se^t)(se^{-t})^3](se^t) + 9(se^t)^2(se^{-t})^2(se^{-t}) \)
\( = [3s^2e^{2t} - 6(s^4e^t e^{-3t})](se^t) + 9(s^4e^{2t}e^{-2t})(se^{-t}) \)
\( = [3s^2e^{2t} - 6s^4e^{-2t}](se^t) + 9s^4(se^{-t}) \)
\( = 3s^3e^{3t} - 6s^5e^{-t} + 9s^5e^{-t} \)
\( = 3s^3e^{3t} + 3s^5e^{-t} \)
In simple words: This problem asks us to find how the function \( z \) changes when \( s \) changes (called partial derivative with respect to \( s \)) and when \( t \) changes (called partial derivative with respect to \( t \)). Since \( z \) depends on \( x \) and \( y \), which in turn depend on \( s \) and \( t \), we used the chain rule. This involves breaking down the total change into smaller parts and then putting them all back together. We carefully calculated each step, paying attention to the exponents and signs.

🎯 Exam Tip: Be meticulous with exponent rules when substituting variables that are themselves exponential functions. Remember that \( e^a \cdot e^b = e^{a+b} \) and \( (e^a)^b = e^{ab} \).

Question 9. W(x, y, z) = xy + yz + zx, x = u – v, y = uv, z = u + v, u, v ∈ R. Find \( \frac{\partial W}{\partial u} \), \( \frac{\partial W}{\partial v} \) evaluate them at \( (\frac{1}{2}, 1) \).
Answer: Given the function \( W(x, y, z) = xy + yz + zx \) and the substitutions \( x = u-v, y = uv, z = u+v \). We need to find \( \frac{\partial W}{\partial u} \) and \( \frac{\partial W}{\partial v} \) and evaluate them at \( (u, v) = (\frac{1}{2}, 1) \).
First, find the partial derivatives of \( W \) with respect to \( x, y, \) and \( z \):
\( \frac{\partial W}{\partial x} = \frac{\partial}{\partial x} (xy + yz + zx) = y + z \)
\( \frac{\partial W}{\partial y} = \frac{\partial}{\partial y} (xy + yz + zx) = x + z \)
\( \frac{\partial W}{\partial z} = \frac{\partial}{\partial z} (xy + yz + zx) = y + x \)
Next, find the partial derivatives of \( x, y, \) and \( z \) with respect to \( u \) and \( v \):
For \( x = u-v \):
\( \frac{\partial x}{\partial u} = 1 \)
\( \frac{\partial x}{\partial v} = -1 \)
For \( y = uv \):
\( \frac{\partial y}{\partial u} = v \)
\( \frac{\partial y}{\partial v} = u \)
For \( z = u+v \):
\( \frac{\partial z}{\partial u} = 1 \)
\( \frac{\partial z}{\partial v} = 1 \)

**1. Find \( \frac{\partial W}{\partial u} \):**
Apply the chain rule for \( \frac{\partial W}{\partial u} \):
\( \frac{\partial W}{\partial u} = \frac{\partial W}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial W}{\partial y} \frac{\partial y}{\partial u} + \frac{\partial W}{\partial z} \frac{\partial z}{\partial u} \)
\( = (y+z)(1) + (x+z)(v) + (y+x)(1) \)
Substitute \( x = u-v, y = uv, z = u+v \):
\( = (uv + (u+v))(1) + ((u-v) + (u+v))(v) + (uv + (u-v))(1) \)
\( = uv + u + v + (2u)v + uv + u - v \)
\( = uv + u + v + 2uv + uv + u - v \)
\( = 4uv + 2u \)
Now, evaluate this at \( (u, v) = (\frac{1}{2}, 1) \):
\( \frac{\partial W}{\partial u} \text{ at } (\frac{1}{2}, 1) = 4(\frac{1}{2})(1) + 2(\frac{1}{2}) \)
\( = 2 + 1 = 3 \)

**2. Find \( \frac{\partial W}{\partial v} \):**
Apply the chain rule for \( \frac{\partial W}{\partial v} \):
\( \frac{\partial W}{\partial v} = \frac{\partial W}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial W}{\partial y} \frac{\partial y}{\partial v} + \frac{\partial W}{\partial z} \frac{\partial z}{\partial v} \)
\( = (y+z)(-1) + (x+z)(u) + (y+x)(1) \)
Substitute \( x = u-v, y = uv, z = u+v \):
\( = -(uv + (u+v)) + ((u-v) + (u+v))(u) + (uv + (u-v))(1) \)
\( = -uv - u - v + (2u)u + uv + u - v \)
\( = -uv - u - v + 2u^2 + uv + u - v \)
\( = 2u^2 - 2v \)
Now, evaluate this at \( (u, v) = (\frac{1}{2}, 1) \):
\( \frac{\partial W}{\partial v} \text{ at } (\frac{1}{2}, 1) = 2(\frac{1}{2})^2 - 2(1) \)
\( = 2(\frac{1}{4}) - 2 \)
\( = \frac{1}{2} - 2 = -\frac{3}{2} \)
In simple words: The function \( W \) depends on \( x, y, z \), and these in turn depend on \( u \) and \( v \). We want to know how \( W \) changes when \( u \) or \( v \) changes. We use the chain rule, which means we find how \( W \) changes with \( x, y, z \), and then how \( x, y, z \) change with \( u \) or \( v \). We combine these changes by multiplying them and adding them up. Finally, we put the given numbers for \( u \) and \( v \) into our results to find the exact rates of change at that specific point.

🎯 Exam Tip: Pay close attention to signs, especially when taking derivatives like \( \frac{\partial x}{\partial v} = -1 \). Carefully group and combine like terms after substitution to simplify the expression before final evaluation.