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Derivative formulas

Monday 30 January 2023

                                             

                             

 DERIVATIVE  FORMULAS

                                        

      Power Rule:                

           

                        (d/dx) (xn ) = nxn-1



    Derivative of a constant

                                       

                                        (d/dx) (a) = 0

   


    Derivative of a constant multiplied with function f(x)

                                               

                       (d/dx) (a. f) a (d/dx) f(x)


    Sum Rule

    (d/dx) [f (x)+ g(x)] =   (d/dx)f (x)+ (d/dx) g(x)

               

    (d/dx) [(f(x) - g(x)] =   (d/dx)f(x) -  (d/dx) g(x)



   Product Rule

(d/dx) [f (x)*g(x)]=  f(x) (d/dx)g(x) + g (x) (d/dx) f (x)


  Quotient Rule

         d/dx (f(x)/g(x))=   g(x) d/dx f(x)-f (x)d/dx g(x)

                                         g(x)²

  Chain Rule
 
                    d/dx f[g(x)] = d/dx f[g(x)] d/dx g(x)


  Derivatives of some trigonometric functions


   (d/dx) sin x = cos x

   (d/dx) cos x = -sin x

   (d/dx) tan x =  sec2 x

   (d/dx) cot x = - cosec2 x

   (d/dx) sec x = sec x  tan x

   (d/dx) cosec x = - cosec x  cot x


Derivatives of elementary functions

  

     (d/dx ) log x = 1/ x

     (d/dx)  ex  = ex 

     (d/dx)  ax = ax. log a , where a>0, a ≠ 1

.a , where a > 0, a ≠ 1      (d/dx) √x =1/(2 √x)

     (d/dx)  log ax   = 1/(log a)x,    x > 0

     (d/dx)  k =0 ,  where k is a constant


Derivatives of hyperbolic trigonometric functions

    

     (d/dx ) sin hx = cos hx

     (d/dx ) cos hx = sin hx

     (d/dx ) tan hx =  sechx

     (d/dx ) cot hx = -cosechx

     (d/dx ) sec hx = -sech x tanh x

     (d/dx ) cosec hx = -cosech x cot hx

     

 Derivatives of Inverse trigonometric functions

     (d/dx )   sin -1 x = 1∕√(1-x²) ,   -1<x<1

     (d/dx  cos -1 x = -1∕√(1-x²) ,    -1<x<1

     (d/dx )  tanh -1 x =  1/1+x²

     (d/dx )  cot -1 x = -1/1+x²

     (d/dx )  cosec -1 x = - 1/|x|√x²-1 ,  |x|>1


Derivatives of Inverse hyperbolic functions

 

      (d/dx ) sinh -1 x = 1/(√x²+1)

      (d/dx ) cosh -1 x = 1/(√x²-1) 

      (d/dx ) tanh -1 x = 1/(1-x²)

      (d/dx ) coth -1 x = 1/x(√1-x²)

      (d/dx ) cosech -1 x = 1/x(√1+x²)

                   


Derivative Formula

Differentiation  or derivative is a process, where we find the instantaneous 

rate of change in function based on one of its variables. The most 

common example is the rate change of displacement with respect to time, 

called velocity. The opposite of finding a derivative is anti-differentiation

Differentiation is a method of finding the  derivative of a function

  If x is a variable and y is another variable, then the rate of change of x with 

respect to y is given by dy/dx. 

This is the general expression of derivative of a function and is represented as 

f'(x) = dy/dx, where y = f(x) is any function.

In real life, the derivatives are used by the people for maintaining a reservoir ,need 

to know when will a reservoir overflow knowing the depth of the water at several 

instances of time, Rocket scientists need to compute the precise velocity with 

which the satellite needs to shot out from the rocket knowing the height of the 

rocket at various times. Financial institutions needs to predict the changes in the 

value of a particular stock knowing its present value. Many such cases it is 

desirable to know how a particular parameter is changing with respect to some 

other parameter.


PROBLEMS USING DERIVATIVE FORMULAS


 For some constants a and b, find the derivative of

(i) (x − a) (x − b)


(ii) (ax2 + b)2


Solution:

(i) (x – a) (x – b)

        

        f(x) = (x-a)(x-b)         

        

        f(x)= x²- (a+ b)x+ ab


          By differentiating on both sides


       d/dx f(x)  = d/dx (x²- (a+ b)x+ ab)


                      =d/dx x²- (a +b)d/dx(x)+ d/dx (ab)

                        

                      =2x - (a+ b) *1 + 0 [ since  (d/dx) (xn ) = nxn-1


                                                                  and      (d/dx) (a) = 0  ]


                    =  2x -a-b



(ii) (ax2 + b)2

              

           f(x) =  (ax2 + b)2   

 Differentiating on both sides by  

                dy/dx f(x)  = d/dx  (ax2 + b)2 


                   d/dx f(x) = a d/dx ( x2 )+ d/dx( b2 )

  

                        = 4ax (ax2 + b)


Find the derivative of


(i) (5x3 + 3x – 1) (x – 1)


(ii) x-3 (5 + 3x)



Solutions

     

     (i) (5x3 + 3x – 1) (x – 1)

      

         f(x) = (5x3 + 3x – 1) (x – 1)

      

         Differentiating on both sides by product rule

        

        d/dx f (x) = d/dx (5x3 + 3x – 1) (x – 1)

             

               =  (5x3 + 3x – 1) d/dx  (x-1) +(x-1) d/dx (5x3 + 3x – 1)

  

              =  (5x3 + 3x – 1) d/dx (x) - d/dx (1) +(x-1) d/dx  (5x3 ) + d/dx (3x )– d/dx (1)


              =  (5x3 + 3x – 1) * 1- 0 +(x-1)(15x2+3)

          

              = 5x3 + 3x – 1+15x3+3x-15x2-3


              = 20 x-15 x2+6x-4


(ii)  x-3 (5 + 3x)

     solution

              f(x) =  x-3 (5 + 3x)

           

             Differentiating on both sides by product rule

            

              d/dx f(x) = d/dx [x-3 (5 + 3x)]

                          

                            = x-3 d/dx(5 + 3x)+(5 + 3x) d/dx  x-3 

                           

                           = x-3 d/dx(5) + d/dx (3x)+(5 + 3x) d/dx  x-3 

                          

                           = x-3  (0+3)+(5+3x) (-3 x-3-1)

                           

                           = 3x-3-15x-4-9x-3



                           = -6 x-3-15x-4

                    

                                                                                                                                                                    

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