Class 12 Maths Revision Notes for Application of Derivatives of Chapter 6


Class 12 Maths Revision Notes for Application of Derivatives of Chapter 6 – Free PDF Download

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CBSE Class 12 Maths Revision Notes Chapter 6 Application of Derivatives

  • If a quantity y varies with another quantity x, satisfying some rule y = f(x), thenApplication of Derivatives Class 12 Notes Mathematics represents the rate of change of y with respect to x and Application of Derivatives Class 12 Notes Mathematicsrepresents the rate of change of y with respect to x at x = xo.
  • If two variables x and y are varying with respect to another variable t, i.e., if   then by Chain Rule
  • A function f is said to be increasing on an interval (a,  b) if  in  for all   Alternatively, if   for each  in, then f(x) is an increasing funciton on (a, b).
  • A function f is said to be decreasing on an interval (a,  b) if  in  for all   Alternatively, if  Application of Derivatives Class 12 Notes Mathematics for each  in, then f(x) is an decreasing funciton on (a, b).
  • The equation of the tangent at    to the curve y =  f (x) is given by
  • If   does not exist at the point Application of Derivatives Class 12 Notes Mathematics, then the tangent at this point is parallel to the y-axis and its equation is.
  • If tangent to a curve  is parallel to x-axis, then = 0
  • Equation of the normal to the curve y = f (x) at a point is given by 
  • If  at the point is zero, then equation of the normal is .
  • If  at the point  does not exist, then the normal is parallel to x-axis and its equation is.
  • Let y = f (x), ∆x be a small increment in x and ∆y be the increment in y corresponding to the increment in x, i.e., ∆y = f (x + ∆x) –  f (x). Then  given by  or Application of Derivatives Class 12 Notes Mathematics is a good appApplication of Derivatives Class 12 Notes Mathematicsroximation of ∆y when  dx x = ∆  is relatively small and we denote it by dy ≈ ∆y.
  • A point c in the domain of a function f at which either f ′(c) = 0 or f is not differentiable is called a critical point of f.
  • First Derivative Test : Let f  be a function defined on an open interval I. Let f   be continuous at a critical point c in I. Then,

(i) If f ′(x) changes sign from positive to negative as x increases through c, i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.
(ii) If f ′(x) changes sign from negative to positive as x increases through c, i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.
(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. In fact, such a point is called point of inflexion.

  • Second Derivative Test: Let f be a function defined on an interval I and c ∈ I. Let f   be twice differentiable at c. Then,

(i) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0
The values f (c) is local maximum value of   f.
(ii) x = c is a point of local minima if f ′(c) = 0 and f ″(c) > 0
In this case, f (c) is local minimum value of f.
(iii) The test fails if f ′(c) = 0 and f ″(c) = 0.
In this case, we go back to the first derivative test and find whether c is a point of maxima, minima or a point of inflexion.

  • Working rule for finding absolute maxima and/or absolute minima

Step 1: Find all critical points of f in the interval, i.e., find points x where either f ′(x) = 0 or f is not differentiable.
Step 2: Take the end points of the interval.
Step 3: At all these points (listed in Step 1 and 2), calculate the values of f.
Step 4: Identify the maximum and minimum values of f out of the values calculated in Step 3.
This maximum value will be the absolute maximum value of f and the minimum value will be the absolute minimum value of f .