Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Differentiation Miscellaneous Exercise 9 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

I. Differentiate the following functions w.r.t.x.

Question 1.
x⁵
Solution:
Let y = x⁵
Differentiating w.r.t. x, we get
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} x^{5}=5 x^{4}$

Question 2.
x^-2
Solution:
Let y = x^-2
Differentiating w.r.t. x, we get
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{-2}\right)=-2 x^{-3}=\frac{-2}{x^{3}}$

Question 3.
√x
Solution:
Let y = √x
Differentiating w.r.t. x, we get
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} \sqrt{x}=\frac{1}{2 \sqrt{x}}$

Question 4.
x√x
Solution:
Let y = x√x
∴ y = $x^{\frac{3}{2}}$
Differentiating w.r.t. x, we get
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{3}{2}}=\frac{3}{2} x^{\frac{1}{2}}$

Question 5.
$\frac{1}{\sqrt{x}}$
Solution:
Let y = $\frac{1}{\sqrt{x}}$
∴ y = $x^{\frac{-1}{2}}$
Differentiating w.r.t. x, we get
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-1}{2} x^{\frac{-3}{2}}=\frac{-1}{2 x^{\frac{3}{2}}}$

Question 6.
7^x
Solution:
Let y = 7^x
Differentiating w.r.t. x, we get
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} 7^{x}=7^{x} \log 7$

II. Find $\frac{d y}{d x}$ if

Question 1.
y = x² + $\frac{1}{x^{2}}$
Solution:

Question 2.
y = (√x + 1)²
Solution:

Question 3.
y = $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}$
Solution:

Question 4.
y = x³ – 2x² + √x + 1
Solution:

Question 5.
y = x² + 2x – 1
Solution:

Question 6.
y = (1 – x)(2 – x)
Solution:

Question 7.
y = $\frac{1+x}{2+x}$
Solution:

Question 8.
y = $\frac{(\log x+1)}{x}$
Solution:

Question 9.
y = $\frac{e^{x}}{\log x}$
Solution:

Question 10.
y = x log x (x² + 1)
Solution:

III. Solve the following:

Question 1.
The relation between price (P) and demand (D) of a cup of Tea is given by D = $\frac{12}{P}$. Find
the rate at which the demand changes when the price is ₹ 2/-. Interpret the result.
Solution:
Demand, D = $\frac{12}{P}$
Rate of change of demand

When price P = 2,
Rate of change of demand,
$\left(\frac{\mathrm{dD}}{\mathrm{dP}}\right)_{\mathrm{P}=2}=\frac{-12}{(2)^{2}}=-3$
∴ When the price is 2, the rate of change of demand is -3.
∴ Here, the rate of change of demand is negative demand would fall when the price becomes ₹ 2.

Question 2.
The demand (D) of biscuits at price P is given by D = $\frac{64}{P^{3}}$, find the marginal demand
when the price is ₹ 4/-.
Solution:
Given demand D = $\frac{64}{P^{3}}$
Now, marginal demand

When P = 4
Marginal demand

Question 3.
The supply S of electric bulbs at price P is given by S = 2p³ + 5. Find the marginal supply when the price is ₹ 5/-. Interpret the result.
Solution:
Given, supply S = 2p³ + 5
Now, marginal supply

∴ When p = 5
Marginal supply = $\left(\frac{\mathrm{dS}}{\mathrm{dp}}\right)_{\mathrm{p}=5}$
= 6(5)²
= 150
Here, the rate of change of supply with respect to the price is positive which indicates that the supply increases.

Question 4.
The total cost of producing x items is given by C = x² + 4x + 4. Find the average cost and the marginal cost. What is the marginal cost when x = 7?
Solution:
Total cost C = x² + 4x + 4
Now. Average cost = $\frac{C}{x}=\frac{x^{2}+4 x+4}{x}$
= x + 4 + $\frac{4}{x}$
and Marginal cost = $\frac{\mathrm{dC}}{\mathrm{d} x}=\frac{\mathrm{d}}{\mathrm{d} x}$(x² + 4x + 4)
= $\frac{\mathrm{d}}{\mathrm{d} x}$ (x²) + 4$\frac{\mathrm{d}}{\mathrm{d} x}$ (x) + $\frac{\mathrm{d}}{\mathrm{d} x}$ (4)
= 2x + 4(1) + 0
= 2x + 4
∴ When x = 7,
Marginal cost = $\left(\frac{\mathrm{d} \mathrm{C}}{\mathrm{d} x}\right)_{x=7}$
= 2(7) + 4
= 14 + 4
= 18

Question 5.
The demand D for a price P is given as D = $\frac{27}{P}$, find the rate of change of demand when the price is ₹ 3/-.
Solution:
Demand, D = $\frac{27}{P}$
Rate of change of demand = $\frac{dD}{dP}$

When price P = 3,
Rate of change of demand,
$\left(\frac{\mathrm{dD}}{\mathrm{dP}}\right)_{\mathrm{P}=3}=\frac{-27}{(3)^{2}}=-3$
∴ When price is 3, Rate of change of demand is -3.

Question 6.
If for a commodity; the price demand relation is given as D = $\left(\frac{P+5}{P-1}\right)$. Find the marginal demand when price is ₹ 2/-
Solution:

Question 7.
The price function P of a commodity is given as P = 20 + D – D² where D is demand. Find the rate at which price (P) is changing when demand D = 3.
Solution:
Given, P = 20 + D – D²
Rate of change of price = $\frac{dP}{dD}$
= $\frac{d}{dD}$(20 + D – D²)
= 0 + 1 – 2D
= 1 – 2D
Rate of change of price at D = 3 is
$\left(\frac{\mathrm{dP}}{\mathrm{dD}}\right)_{\mathrm{D}=3}$ = 1 – 2(3) = -5
∴ Price is changing at a rate of -5, when demand is 3.

Question 8.
If the total cost function is given by C = 5x³ + 2x² + 1; find the average cost and the marginal cost when x = 4.
Solution:
Total cost function C = 5x³ + 2x² + 1
Average cost = $\frac{C}{x}$
= $\frac{5 x^{3}+2 x^{2}+1}{x}$
= 5x² + 2x + $\frac{1}{x}$
When x = 4,
Average cost = 5(4)² + 2(4) + $\frac{1}{4}$
= 80 + 8 + $\frac{1}{4}$
= $\frac{320+32+1}{4}$
= $\frac{353}{4}$
Marginal cost = $\frac{\mathrm{dC}}{\mathrm{d} x}$
= $\frac{d}{dx}$ (5x³ + 2x² + 1)
= 5$\frac{d}{dx}$ (x³) + 2 $\frac{d}{dx}$ (x²) + $\frac{d}{dx}$ (1)
= 5(3x²) + 2(2x) + 0
= 15x² + 4x
When x = 4, marginal cost = $\left(\frac{\mathrm{dC}}{\mathrm{d} x}\right)_{x=4}$
= 15(4)² + 4(4)
= 240 + 16
= 256
∴ The average cost and marginal cost at x = 4 are $\frac{353}{4}$ and 256 respectively.

Question 9.
The supply S for a commodity at price P is given by S = P² + 9P – 2. Find the marginal supply when the price is 7/-.
Solution:
Given, S = P² + 9P – 2

∴ The marginal supply is 23, at P = 7.

Question 10.
The cost of producing x articles is given by C = x² + 15x + 81. Find the average cost and marginal cost functions. Find the marginal cost when x = 10. Find x for which the marginal cost equals the average cost.
Solution:
Given, cost C = x² + 15x + 81

If marginal cost = average cost, then
2x + 15 = x + 15 + $\frac{81}{x}$
∴ x = $\frac{81}{x}$
∴ x² = 81
∴ x = 9 …..[∵ x > 0]