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For the function f(x) to be maximum at x = a

• Necessary condition: f'(a) = 0
• Sufficient condition: f”(a) < 0 (Negative)

• Necessary condition: \(\frac{dC}{dx}\) = 0
• Sufficient condition: \(\frac{d^2C}{dx^2}\) > 0
  Where, C = Production cost, x = Production

• Necessary condition: \(\frac{dR}{dx}\) = 0
• Sufficient condition: \(\frac{d^2R}{dx^2}\) < 0
  Where, R = xp; x = Demand, p = Price

• Necessary condition \(\frac{dP}{dx}\) = 0
• Sufficient condition: \(\frac{d^2P}{dx^2}\) < 0

Where, P = R – C; P = Profit, R = Total revenue, C = Total cost ,

\(\frac{dy}{dx} = 0 OR f'(x) = 0\)

y = f(x). Points at which the maximum and minimum values of f(x) are obtained are called the stationary points of function. Necessary condition for stationary points: f'(x) = 0

If R = Total revenue, C = Total production cost, then profit P = R – C.
The conditions for maximisation of profit are as follows:

• Necessary condition: \(\frac{dP}{dx }\)= 0
• Sufficient condition: \(\frac{d^2P}{dx^2 }\)< 0 (Negative)

x Demand of an item, p = Price per unit of an item, Revenue R = x . p.
The conditions for maximisation of revenue are as follows:

• Necessary condition: \(\frac{dR}{dx}\) = 0
• Sufficient condition: \(\frac{d^2R}{dx^2}\) < 0 (Negative)

x = Units of production,

C = Production cost.

The conditions for minimisation of cost are as follows:

• Necessary condition:\(\frac{ dC}{dx} = 0\)
• Sufficient condition: \(\frac{d^2C}{dx^2} > 0\) (Positive)

The ratio of percentage change in demand and the percentage change in price is called elasticity of demand. Thus,

Elasticity of demand = \(− \frac{Percentage \,change\, in \,demand }{ Percentage \,change \,in \,price }\)

• The demand of an item x is a function of the price of an item p, i.e., x = f(p).
• The relation between demand of an item and its price is inverse. Hence, the negative sign is taken in the formula of elasticity of demand.
• Demand x is a function of price p. Hence, taking the derivative of x with respect to p the elasticity of demand is obtained as follows: Elasticity of demand = −\(\frac{p}x⋅\frac{dx}{dp}\)

Marginal Revenue: The change in total revenue with respect to a small change in demand of an item is called marginal revenue.

• x = Demand of an item, p = Price of an item, then total revenue R = x . p
• Marginal revenue is obtained by finding derivative of R with respect to x.

Thus, marginal revenue = \(\frac{dR}{dx}\)

Marginal Cost:

• The change in total cost with respect to a small change in production is called marginal cost.
• x = Production of an item, C = Production cost, then finding derivative of C with respect to x, marginal cost is obtained.

Thus, marginal cost =\(\frac{ dC}{dx}\)

Maximum Value of a Function:
y = f(x). If h is a very small positive number and f(a) > f(a + h) and f(a) > f(a – h), then at x = a, the function f(x) is called maximum.

The conditions for maximisation of function at x = a:

• Necessary condition: f(a) = 0
• Sufficient condition: f”(a) < 0 (Negative)

Minimum Value of a Function:
y = f(x). If h is . a very small positive number and f(a) < f(a + h) and f(a) < f(a – h), then at x = a, the function f(x) is called minimum.

The condition for minimisation of function at x = a:

• Necessary condition: f(a) = 0
• Sufficient condition: f(a) > 0 (Positive)

1. Increasing Function: y = f(x). If h is a very small positive number and f(a + h) >f(a) and f(a)<f(a-h), then at x = a, the function f(x) is called increasing function. That is, if at x = a, the function f(x) is increasing function then f(a) > 0.

2. Decreasing Function: y = fx). If h is a very small positive number and f(a + h) < f(a) and f(a) < f(a-h), then at x = a, the function f(x) is called decreasing function. That is, if at x = a, the function fix) is decreasing function then f(a) < 0.

f(x) = x + 1

⇒ (fof)(x) = f(x) + 1

= (x + 1) + 1

= x + 2

So, \(\frac{d}{d\text x}\)(fof)(x) = \(\frac{d}{d\text x}\)(x + 2)

= 1

We know that  sin^-1x+ cos^-1x = \(\cfrac{\pi}2\)

So, here y = sin^–1 x + cos^–1 x

= \(\cfrac{\pi}2\) which is a constant.

Also, sin^-1 x and cos^-1 x exist only when -1 ≤ x ≤ 1

So, \(\cfrac{dy}{d\mathrm x}\) = 0 when x ∈ [-1, 1] and does not exist for all other values of x.

f(x) is an even function.

This means that f(-x) = f(x).

If we differentiate this equation on both sides w.r.t. x, we get –

f’(-x).(-1) = f’(x)

or, -f’(-x) = f’(x)

i.e., f’(x) is an odd function.

f(x) is an odd function.

This means that f(-x) = -f(x).

If we differentiate this equation on both sides w.r.t. x, we get –

f’(-x).(-1) = -f’(x)

or, f’(-x) = f’(x)

i.e., f’(x) is an even function.

By Product Rule: (uv)′ = u′v + uv′

Here u = x⁴ and v = tan x

(x⁴ tan x)’ = 4x³ × tan x + x⁴ × sec²x

= 4x³tan x + x⁴sec 2x

By Product Rule: (uv)′ = u′v + uv′

Here u = (x² + 3x + 1) and v = sin x

[(x² + 3x + 1) sin x]’ = (2x + 3) × sinx + (x² + 3x + 1) × cosx

= (2x + 3)sin x + (x² + 3x + 1)cos x

By Product Rule: (uv)′ = u′v + uv′

Here u = x³ and v = sec x

(x³ sec x)’ = 3x²(sec x) + x³(sec x tan x)

= x²sec x(3 + x tan x)

By Product Rule: (uv)′ = u′v + uv′

Here u = xⁿ and v = cot x

(xⁿ cot x)’ = nx^n-1 (cot x) + xⁿ (- cosec²x)

= nx^n-1 (cot x) – xⁿ (cosec²x)

= x^n-1 (n cot x – x cosec²x)

By Product Rule: (uv)′ = u′v + uv′

Here u = e^x and v = cot x

(e^x cot x)’ = e^x (cot x) + e^x (- cosec²x)

= e^x (cot x) – e^x cosec²x

= e^x (cot x – cosec²x)