Revision: Mathematics >> Applications of Derivatives CUET (UG) Applications of Derivatives

A function ( f(x) ) is said to be a decreasing function on (a, b) if x₁ < x₂ ⇒ f(x₁) ≥ f(x₂)

Strictly Decreasing Function:

• If x₁ < x₂ ⇒ f(x₁) > f(x₂)

A function ( f ) is said to be monotonic in an interval if it is either increasing or decreasing in that interval.

A function ( f(x) ) is said to be an increasing function on ((a, b)) if x₁ < x₂ ⇒ f(x₁) ≤ f(x₂)

Strictly Increasing Function:

• If x₁ < x₂ ⇒ f(x₁) < f(x₂)

\[\mathrm{f(a+h)\approx f(a)+h~f^{\prime}(a)}\]

Given:

y `sqrt(x^2 + 1) = log (sqrt(x^2 + 1) - x)`

Differentiate the Left-Hand Side:

Using the product rule (uv)′ = u′v + uv′:

Let u = y and v = `sqrt(x^2 + 1)`

`d/dx (y sqrt(x^2 + 1)) = (dy)/(dx) . sqrt(x^2 + 1) + y . d/dx (sqrt(x^2 + 1))`

= `sqrt(x^2 + 1) (dy)/(dx)  + y . (1/(2sqrt(x^2 + 1)) . 2x)`

= `sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1)`   ...(i)

Differentiate the Right-Hand Side:

Using the chain rule for log(u):

`d/dx [log (sqrt(x^2 + 1) - x)] = 1/(sqrt(x^2 + 1) - x) . d/dx (sqrt(x^2 + 1) - x)`

= `1/(sqrt(x^2 + 1) - x) . (x/sqrt(x^2 + 1) - 1)`

Take the LCM in the bracket:

= `1/(sqrt(x^2 + 1) - x) . ((x - sqrt(x^2 + 1))/sqrt(x^2 + 1))`

= `1/(sqrt(x^2 + 1) - x) . ((-sqrt(x^2 + 1) - x)/sqrt(x^2 + 1))`

= `-1/(sqrt(x^2 + 1)`   ...(ii)

Equate LHS and RHS

`sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1) = -1/(sqrt(x^2 + 1)`

Multiply the entire equation by `sqrt(x^2 + 1)` to clear the denominators:

`(sqrt(x^2 + 1) . sqrt(x^2 + 1)) (dy)/(dx) + xy = -1`

`(x^2 + 1) (dy)/(dx) + xy = -1`

`(x^2 + 1) (dy)/(dx) + xy + 1 = 0`

Hence proved

• f′(x) > 0 ⇒ function increasing
• f′(x) < 0 ⇒ function decreasing
• f′(x) = 0 ⇒ function constant

First Derivative Test:

Maxima at x = c:

• f′(c) = 0
• f′(c − h) > 0 and f′(c + h) < 0

Minima at x = c:

• f′(c) = 0
• f′(c − h) < 0 and f′(c + h) > 0

Second Derivative Test:

• If f′(a) = 0 and f″(a) < 0 → Maximum
• If f′(a) = 0 and f″(a) > 0 → Minimum
• If f″(a) = 0 → Test fails (use first derivative)