Maharashtra Board Class 12 Maths Part 1 Chapter 3 Differentiation 3.5 Solutions

1. Find dy/dx if:

Question 1. x = at², y = 2at
Answer:
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get
\( \frac{dx}{dt} = \frac{d}{dt}(at^2) = a \times 2t = 2at \)
and \( \frac{dy}{dt} = \frac{d}{dt}(2at) = 2a \times 1 = 2a \)
\( \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{2a}{2at} = \frac{1}{t} \)
In simple words: To find dy/dx, we first differentiate x and y with respect to the parameter 't' separately, then divide dy/dt by dx/dt.

🎯 Exam Tip: Remember to use the chain rule for parametric differentiation. Clearly show all differentiation steps for each variable with respect to the parameter.

Question 2. x = 2at², y = at⁴
Answer:
Solution:
x = 2at²
Differentiating both sides w.r.t. t, we get
\( \frac{dx}{dt} = 4at \)
y = at⁴
Differentiating both sides w.r.t. t, we get
\( \frac{dy}{dt} = 4at^3 \)
\( \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{4at^3}{4at} = t^2 \)
In simple words: We find the derivatives of x and y concerning 't', then combine them using the formula dy/dx = (dy/dt) / (dx/dt) to get the final derivative.

🎯 Exam Tip: Pay attention to the power rule when differentiating terms involving 't' to different powers. Simplify the final expression carefully.

Question 3. x = e³ᵗ, y = e⁽⁴ᵗ⁺⁵⁾
Answer:
Solution:
x = e³ᵗ, y = e⁽⁴ᵗ⁺⁵⁾
Differentiating x and y w.r.t. t, we get
\( \frac{dx}{dt} = \frac{d}{dt}(e^{3t}) = e^{3t} \cdot \frac{d}{dt}(3t) \)
\( = e^{3t} \times 3 \times 1 = 3e^{3t} \)
and \( \frac{dy}{dt} = \frac{d}{dt}[e^{(4t+5)}] = e^{(4t+5)} \cdot \frac{d}{dt}(4t+5) \)
\( = e^{(4t+5)} \times (4 \times 1 + 0) = 4e^{(4t+5)} \)
\( \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{4e^{(4t+5)}}{3e^{3t}} \)
\( = \frac{4}{3} e^{4t+5-3t} = \frac{4}{3} e^{t+5} \)
In simple words: For exponential functions, differentiate the exponent first, then multiply by the original exponential function. Then, divide the y-derivative by the x-derivative.

🎯 Exam Tip: Remember the chain rule for exponential functions \( \frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x) \). Simplify the exponential terms using exponent rules.

2. Find dy/dx if:

Question 1. \( x = (u + \frac{1}{u})^2 \), \( y = (2)^{(u+\frac{1}{u})} \)
Answer:
Solution:
\( x = (u + \frac{1}{u})^2 \), \( y = (2)^{(u+\frac{1}{u})} \) ......(1)
Differentiating x and y w.r.t. u, we get,
\( \frac{dx}{du} = \frac{d}{du}(u + \frac{1}{u})^2 = 2(u + \frac{1}{u}) \frac{d}{du}(u + \frac{1}{u}) \)
\( = 2(u + \frac{1}{u})(1 - \frac{1}{u^2}) \)
and \( \frac{dy}{du} = \frac{d}{du}[2^{(u+\frac{1}{u})}] \)
\( = 2^{(u+\frac{1}{u})} \cdot \log 2 \cdot \frac{d}{du}(u + \frac{1}{u}) \)
\( = 2^{(u+\frac{1}{u})} \cdot \log 2 \cdot (1 - \frac{1}{u^2}) \)
\( \frac{dy}{dx} = \frac{(dy/du)}{(dx/du)} \)
\( = \frac{2^{(u+\frac{1}{u})} \cdot \log 2 \cdot (1 - \frac{1}{u^2})}{2(u + \frac{1}{u})(1 - \frac{1}{u^2})} \)
\( = \frac{2^{(u+\frac{1}{u})} \cdot \log 2}{2(u + \frac{1}{u})} \)
From (1), \( y = 2^{(u+\frac{1}{u})} \) and \( \sqrt{x} = u + \frac{1}{u} \)
\( \frac{dy}{dx} = \frac{y \log 2}{2\sqrt{x}} \) ... [By (1)]
In simple words: This problem involves differentiating parametric equations where the parameter is 'u'. We find dx/du and dy/du using standard differentiation rules, then divide dy/du by dx/du to find dy/dx. Finally, substitute back terms of x and y.

🎯 Exam Tip: Be careful with the differentiation of terms like \( a^{f(x)} \) which is \( a^{f(x)} \log a \cdot f'(x) \) and the power rule. Algebraic simplification after division is key.

Question 2. \( x = \sqrt{1 + u^2} \), \( y = \log(1 + u^2) \)
Answer:
Solution:
\( x = \sqrt{1 + u^2} \), \( y = \log(1 + u^2) \) ......(1)
Differentiating x and y w.r.t. u, we get,
\( \frac{dx}{du} = \frac{d}{du}(\sqrt{1 + u^2}) = \frac{1}{2\sqrt{1 + u^2}} \cdot \frac{d}{du}(1 + u^2) \)
\( = \frac{1}{2\sqrt{1 + u^2}} \times (0 + 2u) = \frac{u}{\sqrt{1 + u^2}} \)
and \( \frac{dy}{du} = \frac{d}{du}[\log(1 + u^2)] \)
\( = \frac{1}{1 + u^2} \cdot \frac{d}{du}(1 + u^2) \)
\( = \frac{1}{1 + u^2} \times (0 + 2u) = \frac{2u}{1 + u^2} \)
\( \frac{dy}{dx} = \frac{(dy/du)}{(dx/du)} \)
\( = \frac{(\frac{2u}{1 + u^2})}{(\frac{u}{\sqrt{1 + u^2}})} \)
\( = \frac{2u}{1 + u^2} \times \frac{\sqrt{1 + u^2}}{u} \)
\( = \frac{2}{\sqrt{1 + u^2}} \)
In simple words: We find the derivatives of x and y with respect to 'u'. Then, we divide the derivative of y by the derivative of x and simplify the resulting expression.

🎯 Exam Tip: Remember the chain rule for square roots \( \frac{d}{dx}(\sqrt{f(x)}) = \frac{f'(x)}{2\sqrt{f(x)}} \) and for logarithmic functions \( \frac{d}{dx}(\log f(x)) = \frac{f'(x)}{f(x)} \). Simplify fractions carefully.

Question 3. Differentiate 5ˣ with respect to log x.
Answer:
Solution:
Let u = 5ˣ and v = log x
Then we want to find \( \frac{du}{dv} \)
Differentiating u and v w.r.t. x, we get
\( \frac{du}{dx} = \frac{d}{dx}(5^x) = 5^x \cdot \log 5 \)
and \( \frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x} \)
\( \frac{du}{dv} = \frac{(du/dx)}{(dv/dx)} = \frac{5^x \cdot \log 5}{(\frac{1}{x})} \)
\( = x \cdot 5^x \cdot \log 5 \)
In simple words: To differentiate one function with respect to another, we treat them as parametric equations with 'x' as the parameter. We find their derivatives with respect to x, then divide.

🎯 Exam Tip: Recall the derivative of \( a^x \) is \( a^x \log a \) and the derivative of \( \log x \) is \( 1/x \). Ensure correct application of these formulas and proper fraction division.

3. Solve the following:

Question 1. If \( x = a (1 - \frac{1}{t}) \), \( y = a (1 + \frac{1}{t}) \), then show that \( \frac{dy}{dx} = -1 \)
Answer:
Solution:
\( x = a(1 - \frac{1}{t}) \), \( y = a(1 + \frac{1}{t}) \)
Differentiating x and y w.r.t. t, we get
\( \frac{dx}{dt} = a \frac{d}{dt}(1 - t^{-1}) = a[0 - (-1)t^{-2}] = \frac{a}{t^2} \)
and \( \frac{dy}{dt} = a \frac{d}{dt}(1 + t^{-1}) = a[0 + (-1)t^{-2}] = -\frac{a}{t^2} \)
\( \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{(-\frac{a}{t^2})}{(\frac{a}{t^2})} = -1 \)
In simple words: We differentiate both x and y with respect to 't', treating \( \frac{1}{t} \) as \( t^{-1} \). Then, we divide the derivative of y by the derivative of x, which simplifies to -1.

🎯 Exam Tip: Convert reciprocal terms to negative powers for easier differentiation. Pay close attention to signs during differentiation and final simplification.

Question 2. If \( x = \frac{4t}{1+t^2} \), \( y = 3 \frac{1-t^2}{1+t^2} \), then show that \( \frac{dy}{dx} = -\frac{9x}{4y} \)
Answer:
Solution:
\( x = \frac{4t}{1+t^2} \), \( y = 3 \frac{1-t^2}{1+t^2} \)
Differentiating x and y w.r.t. t, we get
\( \frac{dx}{dt} = \frac{d}{dt}(\frac{4t}{1+t^2}) \)
\( = \frac{(1+t^2)\frac{d}{dt}(4t) - 4t \frac{d}{dt}(1+t^2)}{(1+t^2)^2} \)
\( = \frac{(1+t^2)(4) - 4t(0+2t)}{(1+t^2)^2} \)
\( = \frac{4+4t^2 - 8t^2}{(1+t^2)^2} = \frac{4-4t^2}{(1+t^2)^2} = \frac{4(1-t^2)}{(1+t^2)^2} \)
and \( \frac{dy}{dt} = 3 \frac{d}{dt}(\frac{1-t^2}{1+t^2}) \)
\( = 3 \frac{(1+t^2)\frac{d}{dt}(1-t^2) - (1-t^2)\frac{d}{dt}(1+t^2)}{(1+t^2)^2} \)
\( = 3 \frac{(1+t^2)(-2t) - (1-t^2)(2t)}{(1+t^2)^2} \)
\( = 3 \frac{-2t-2t^3 - 2t+2t^3}{(1+t^2)^2} \)
\( = \frac{3(-4t)}{(1+t^2)^2} = \frac{-12t}{(1+t^2)^2} \)
\( \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} \)
\( = \frac{(\frac{-12t}{(1+t^2)^2})}{(\frac{4(1-t^2)}{(1+t^2)^2})} \)
\( = \frac{-12t}{4(1-t^2)} = \frac{-3t}{1-t^2} \) ......(1)
Now, we need to express this in terms of x and y.
We have \( x = \frac{4t}{1+t^2} \) and \( y = 3 \frac{1-t^2}{1+t^2} \)
Consider \( -\frac{9x}{4y} = -\frac{9(\frac{4t}{1+t^2})}{4(3 \frac{1-t^2}{1+t^2})} \)
\( = -\frac{9 \times 4t}{(1+t^2)} \times \frac{(1+t^2)}{4 \times 3 (1-t^2)} \)
\( = -\frac{36t}{12(1-t^2)} = \frac{-3t}{1-t^2} \) ......(2)
From (1) and (2)
\( \frac{dy}{dx} = -\frac{9x}{4y} \)
In simple words: This problem uses the quotient rule to differentiate x and y with respect to 't'. After finding dx/dt and dy/dt, we divide them to get dy/dx. Finally, we manipulate the given expressions for x and y to show that dy/dx is equivalent to -9x/(4y).

🎯 Exam Tip: The quotient rule \( \frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2} \) is crucial here. Be meticulous with algebraic simplification, especially when substituting back x and y to match the required proof.

Question 3. If x = t . log t, y = tᵗ, then show that \( \frac{dy}{dx} - y = 0 \).
Answer:
Solution:
x = t log t
Differentiating w.r.t. t, we get
\( \frac{dx}{dt} = \frac{d}{dt}(t \cdot \log t) \)
\( = t \frac{d}{dt}(\log t) + (\log t) \frac{d}{dt}(t) \)
\( = t \times \frac{1}{t} + (\log t) \times 1 = 1 + \log t \)
Also, y = tᵗ
\( \log y = \log t^t = t \log t \)
Differentiating both sides w.r.t. t, we get
\( \frac{1}{y} \frac{dy}{dt} = \frac{d}{dt}(t \log t) \)
\( = t \frac{d}{dt}(\log t) + (\log t) \frac{d}{dt}(t) \)
\( = t \times \frac{1}{t} + (\log t) \times 1 \)
\( = 1 + \log t \)
\( \frac{dy}{dt} = y(1 + \log t) \)
\( \frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)} = \frac{y(1 + \log t)}{1 + \log t} \)
\( \frac{dy}{dx} = y \)
\( \implies \frac{dy}{dx} - y = 0 \)
In simple words: We differentiate x using the product rule. For y, we take the logarithm of both sides to simplify differentiation, then use the product rule. Finally, we divide dy/dt by dx/dt, which results in dy/dx = y, leading to the required proof.

🎯 Exam Tip: When differentiating functions like \( f(x)^{g(x)} \), use logarithmic differentiation. Remember the product rule and chain rule for each step, and ensure careful substitution in the final expression.