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.

x^{5}

Solution:

Let y = x^{5}

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 = x^{2} + \(\frac{1}{x^{2}}\)

Solution:

Question 2.

y = (√x + 1)^{2}

Solution:

Question 3.

y = \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\)

Solution:

Question 4.

y = x^{3} – 2x^{2} + √x + 1

Solution:

Question 5.

y = x^{2} + 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 (x^{2} + 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 = 2p^{3} + 5. Find the marginal supply when the price is ₹ 5/-. Interpret the result.

Solution:

Given, supply S = 2p^{3} + 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 = x^{2} + 4x + 4. Find the average cost and the marginal cost. What is the marginal cost when x = 7?

Solution:

Total cost C = x^{2} + 4x + 4

Now. Average cost = \(\frac{C}{x}=\frac{x^{2}+4 x+4}{x}\)

= x + 4 + \(\frac{4}{x}\)

and Marginal cost = \(\frac{\mathrm{dC}}{\mathrm{d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\)(x^{2} + 4x + 4)

= \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x^{2}) + 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 – D^{2} where D is demand. Find the rate at which price (P) is changing when demand D = 3.

Solution:

Given, P = 20 + D – D^{2}

Rate of change of price = \(\frac{dP}{dD}\)

= \(\frac{d}{dD}\)(20 + D – D^{2})

= 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 = 5x^{3} + 2x^{2} + 1; find the average cost and the marginal cost when x = 4.

Solution:

Total cost function C = 5x^{3} + 2x^{2} + 1

Average cost = \(\frac{C}{x}\)

= \(\frac{5 x^{3}+2 x^{2}+1}{x}\)

= 5x^{2} + 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 = \(\frac{\mathrm{dC}}{\mathrm{d} x}\)

= \(\frac{d}{dx}\) (5x^{3} + 2x^{2} + 1)

= 5\(\frac{d}{dx}\) (x^{3}) + 2 \(\frac{d}{dx}\) (x^{2}) + \(\frac{d}{dx}\) (1)

= 5(3x^{2}) + 2(2x) + 0

= 15x^{2} + 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 = P^{2} + 9P – 2. Find the marginal supply when the price is 7/-.

Solution:

Given, S = P^{2} + 9P – 2

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

Question 10.

The cost of producing x articles is given by C = x^{2} + 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 = x^{2} + 15x + 81

If marginal cost = average cost, then

2x + 15 = x + 15 + \(\frac{81}{x}\)

∴ x = \(\frac{81}{x}\)

∴ x^{2} = 81

∴ x = 9 …..[∵ x > 0]