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Unit-3

Derivatives and applications of derivatives

Q1) Evaluate the

A1)

We can simply find the solution as follows,

Q2) evaluate

A2)

Q3) Find dy/dx of the following functions-

A3)

Let y =

Then-

and

Let y =

Then-

Q4) Differentiate with respect to x.

A4)

Let

Now

Q5) if y = then find dy/dx.

A5)

Suppose z =

Now-

So that-

Q6) If y = log log log then find dy/dx.

A6)

Suppose y = log u where u = log v and v = log

So that-

Q7) if y =

A7)

Let y = log u where u =

Now

Q8) Calculate and for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

A8)

To calculate treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 9x² - 0 + 2y – 8 + 0 – 0

= 9x² + 2y – 8

Similarly partial derivative of f(x,y) with respect to y is:

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 0 – 10y + 2x – 0 + 4 – 0

= 2x – 10y +4.

Q9) if , then show that-

A9)

Here we have,

u = …………………..(1)

now partially differentiate eq.(1) w.r to x and y , we get

………………..(2)

And now,

………………….(3)

Adding eq. (1) and (3) , we get

= 0

Hence proved.

Q10) if y = log x/ x a then find

A10)

First we will find the first derivative-

Now

Q11) Examine for maximum and minimum for the function f(x) =

A11)

Here the first derivative is-

So that, we get-

Now we will get to know that the function is maximum or minimum at these values of x.

For x = 3

Let us assign to x, the values of 3 – h and 3 + h (here h is very small) and put these values at f(x).

Then-

which is negative for h is very small

which is positive.

Thus f’(x) changes sign from negative to positive as it passes through x = 3.

So that f(x) is minimum at x = 3 and the minimum value is-

And f(x) is maximum at x = -3.

Q12) Find out the value of x for which the function f(x) = is minimum of maximum.

A12)

Let y = f(x) =

Then the first derivative will be-

Putting , we get-

By using middle term splitting-

We get-

Now we will find the second derivative-

We get-

Here

At x = 4,

At x = 16,

Q13) The total cost function of the firm is given by-

Where x is the output.

Find out-

Fixed cost

Average cost

Marginal cost

Marginal average cost

A13)

Here we have-

Now

Fixed cost by putting x = 0, we get-

FC = 5000

2. Average cost = TC/x =

3. MC = d TC / dx

4. Marginal average cost = d AC / dx =

Q14) A manufacturer can sell x items per month at a price p = 300 – 2x rupees. Produced items cost the manufacturer y rupees y = 2x + 1000. How much profit will yield maximum profits

A14)

Profit (P) = Sale – total cost = x. p – y

= x (300 – 2x) – (2x + 1000) = 298x – 2 – 1000

For maximum profit-

Again the second derivative is-

Hence the profit will be maximum for 74 items.

Q15) A belt manufacturer produces high grade and low grade shoes in x units (hundred pairs) and y units (hundred pair) per day respectively.

Assume that the equipment is capable of producing the belts in the following relationship-

If the high grade belt are sold at a price twice the low grade belts.

Determine the pairs of both the belts he should produce to maximize his total revenue.

A15)

let the price of low grade belt be Rs. p per pair, then the price of high grade belts will be rs. 2 p per pair.

The total revenue will be-

For maximizing TR(x), take its first derivative w.r.t. x, put it equal to zero and solve for x,

Now put it equals to zero and solving for x,

So that,

x = 8 [ x< 10, and x = 12 is not possible]

Now at x = 8,

RT(x) is maximum at x = 8

Put this value in the function-

So that we can conclude that the total revenue is maximum when the manufacturer produces 800 pairs of high grade and 100 pairs of low grade.

Q16) The Namkeen industry decides to reduce the price of its product, from Rs.100 to Rs. 75. The company expects that the sales of Namkeen will increase from 10,000 units a month to 20,000 units a month. Calculate and comment on the price elasticity of demand.

A16)

First, we need to calculate the percentage change in quantity demanded and percentage change in price. So,

% Change in Price = (Rs. 75-100)/(Rs.100) = -25%

% Change in Demand = (20,000-10,000)/(10,000) = 100%

Therefore, the Price Elasticity of Demand = 100%/-25% = -4.

This means the demand is relatively elastic

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