Maharashtra Board Class 12 Maths Part 1 Chapter 3 Differentiation 3.2 Solutions

Question 1. Find the rate of change of demand (x) of a commodity with respect to price (y) if:
Question 1. y = 12 + 10x + 25x²
Answer:
Solution:
Given \( y = 12 + 10x + 25x^2 \)
\( \frac{dy}{dx} = \frac{d}{dx}(12+ 10x + 25x^2) \)
\( = \frac{d}{dx}(12)+10\frac{d}{dx}(x)+25\frac{d}{dx}(x^2) \)
\( = 0+10 \times 1 +25 \times 2x \)
\( = 10+50x \)
By derivative of inverse function,
\( \frac{dx}{dy} = \frac{1}{(\frac{dy}{dx})} = \frac{1}{10+50x} \)
Hence, the rate of change of demand (x) with respect to price (y) \( = \frac{dx}{dy} = \frac{1}{10+50x} \)
In simple words: To find the rate of change of demand (x) with respect to price (y), we first differentiate y with respect to x, and then use the inverse function rule to find dx/dy. This tells us how demand changes for a given change in price.

🎯 Exam Tip: Remember to apply the chain rule correctly when differentiating terms like \(x^2\) and to use the inverse function derivative rule (\(dx/dy = 1/(dy/dx)\)) when finding the rate of change of demand with respect to price.

Question 2. y = 18x + log(x - 4)
Answer:
Solution:
Given \( y = 18x + \log (x - 4) \)
\( \frac{dy}{dx} = \frac{d}{dx}[18x+\log (x - 4)] \)
\( = 18\frac{d}{dx}(x)+\frac{d}{dx}[\log (x-4)] \)
\( = 18 \times 1 + \frac{1}{x-4}\frac{d}{dx}(x-4) \)
\( = 18+ \frac{1}{x-4}(1-0) \)
\( = 18+ \frac{1}{x-4} = \frac{18(x-4)+1}{x-4} \)
\( = \frac{18x-72+1}{x-4} \)
\( = \frac{18x-71}{x-4} \)
By derivative of inverse function
\( \frac{dx}{dy} = \frac{1}{(\frac{dy}{dx})} = \frac{x-4}{18x-71} \)
Hence, the rate of change of demand (x) with respect to price (y) \( = \frac{dx}{dy} = \frac{x-4}{18x-71} \)
In simple words: We differentiate the given function y with respect to x, taking care of the logarithmic term. Then, we use the inverse derivative formula to find dx/dy, which represents how much demand (x) changes for a unit change in price (y).

🎯 Exam Tip: Pay close attention to the derivative of logarithmic functions (\(\frac{d}{dx} \log(f(x)) = \frac{f'(x)}{f(x)}\)) and ensure algebraic simplification is accurate before applying the inverse derivative rule.

Question 3. y = 25x + log(1 + x²)
Answer:
Solution:
Given \( y = 25x + \log(1 + x^2) \)
\( \frac{dy}{dx} = \frac{d}{dx}[25x+\log (1+x^2)] \)
\( = 25\frac{d}{dx}(x)+\frac{d}{dx}[\log (1+x^2)] \)
\( = 25 \times 1 + \frac{1}{1+x^2}\frac{d}{dx}(1+x^2) \)
\( = 25 + \frac{1}{1+x^2} \times (0+2x) \)
\( = 25 + \frac{2x}{1+x^2} = \frac{25(1+x^2)+2x}{1+x^2} \)
\( = \frac{25+25x^2+2x}{1+x^2} \)
\( = \frac{25x^2+2x+25}{1+x^2} \)
By derivative of inverse function,
\( \frac{dx}{dy} = \frac{1}{(\frac{dy}{dx})} = \frac{1+x^2}{25x^2+2x+25} \)
Hence, the rate of change of demand (x) with respect to price (y) \( \frac{dx}{dy} = \frac{1+x^2}{25x^2+2x+25} \)
In simple words: We differentiate the given function y with respect to x, carefully handling the derivative of the logarithmic term involving \(x^2\). Then, we invert the result to find dx/dy, representing the responsiveness of demand to price changes.

🎯 Exam Tip: Always simplify the expression for \(dy/dx\) as much as possible before taking its reciprocal to find \(dx/dy\). This reduces potential errors in the final answer.

Question 2. Find the marginal demand of a commodity where demand is x and price is y.
Question 1. y = xe⁻ˣ + 7
Answer:
Solution:
Given: \( y = xe^{-x}+7 \)
\( \frac{dy}{dx} = \frac{d}{dx}(xe^{-x}+7) \)
\( = \frac{d}{dx}(xe^{-x})+\frac{d}{dx}(7) \)
\( = x\frac{d}{dx}(e^{-x})+e^{-x}\frac{d}{dx}(x)+0 \)
\( = x \times e^{-x}(-x)+e^{-x} \times 1 \)
\( = xe^{-x}(-1)+e^{-x} \)
\( = e^{-x}(-x+1) \)
\( = \frac{1-x}{e^x} \)
By the derivative of inverse function,
\( \frac{dx}{dy} = \frac{1}{(\frac{dy}{dx})} = \frac{e^x}{1-x} \)
Hence, marginal demand \( \frac{dx}{dy} = \frac{e^x}{1-x} \)
In simple words: Marginal demand is found by differentiating the demand function (x) with respect to price (y). Here, we first differentiated price (y) with respect to demand (x) using the product rule and then inverted the result.

🎯 Exam Tip: When dealing with products of functions like \(xe^{-x}\), remember to correctly apply the product rule for differentiation. Marginal demand is synonymous with \(dx/dy\).

Question 2. y = \(\frac{x+2}{x^2+1}\)
Answer:
Solution:
Given: \( y = \frac{x+2}{x^2+1} \)
\( \frac{dy}{dx} = \frac{d}{dx}(\frac{x+2}{x^2+1}) \)
\( = \frac{(x^2+1)\frac{d}{dx}(x+2)-(x+2)\frac{d}{dx}(x^2+1)}{(x^2+1)^2} \)
\( = \frac{(x^2+1)(1+0)-(x+2)(2x+0)}{(x^2+1)^2} \)
\( = \frac{x^2+1-2x^2-4x}{(x^2+1)^2} \)
\( = \frac{1-x^2-4x}{(x^2+1)^2} \)
By the derivative of inverse function,
\( \frac{dx}{dy} = \frac{1}{(\frac{dy}{dx})} = \frac{(x^2+1)^2}{1-4x-x^2} \)
Hence, marginal demand \( \frac{dx}{dy} = \frac{(x^2+1)^2}{1-4x-x^2} \)
In simple words: We find the marginal demand by first differentiating the price function y with respect to x using the quotient rule. After simplifying, we take the reciprocal of this result to get dx/dy, which is the marginal demand.

🎯 Exam Tip: The quotient rule (\(\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}\)) is essential for these problems. Ensure all terms are correctly differentiated and signs are managed carefully during simplification.

Question 3. y = \(\frac{5x+9}{2x-10}\)
Answer:
Solution:
Given: \( y=\frac{5x+9}{2x-10} \)
\( \frac{dy}{dx} = \frac{d}{dx}(\frac{5x+9}{2x-10}) \)
\( = \frac{(2x-10)\frac{d}{dx}(5x+9)-(5x+9)\frac{d}{dx}(2x-10)}{(2x-10)^2} \)
\( = \frac{(2x-10)(5 \times 1+0)-(5x+9)(2 \times 1-0)}{(2x-10)^2} \)
\( = \frac{10x-50-(10x+18)}{(2x-10)^2} \)
\( = \frac{10x-50-10x-18}{(2x-10)^2} \)
\( = \frac{-68}{(2x-10)^2} \)
By the derivative of inverse function,
\( \frac{dx}{dy} = \frac{1}{(\frac{dy}{dx})} = -\frac{(2x-10)^2}{68} \)
Hence, marginal demand \( \frac{dx}{dy} = -\frac{(2x-10)^2}{68} \)
In simple words: To determine the marginal demand, we apply the quotient rule to differentiate the given price function y with respect to x. After simplifying the derivative, we take its reciprocal to arrive at dx/dy, which quantifies the change in demand per unit change in price.

🎯 Exam Tip: Be meticulous with algebraic simplification after applying the quotient rule, especially when dealing with negative signs. A common error is mismanaging signs when distributing terms.