DIFFERENTIATION OF VARIABLE POWER FUNCTIONS USING PRODUCT RULE

Let us learn and remember  easiest method of differentiation, differentiate,differentiation formula of the functions which involve variable power in it in an easy and short cut manners.

Till Date we have learn to differentiate this question by taking log on both sides and then differentiate.

This is very long process to differentiate this types of functions .
But today we shall learn a different and an easiest method to differentiate such type of functions.
      

then assume this function as [ log (Base) ]× [ Power ] then use product rule of differentiation and place the given function in front of the result so obtained.

f '(x) = [ log (Base) ] d/dx [ Power ] + [ Power ] d/dx [ log (Base) ]

Question  Differentiate f(x) = (cos x )^{sin x}

Consider    h(x) = (log cos x) ×(sin x)

then it derivative will be 

 f '(x) = f(x) Diff  (h(x))

⇒ f '(x) = f(x) [(log cos x) . Diff (sin x) + sin x Diff (log cos x)]

Therefore  f '(x) = f(x) [(log cos x) . cos x + sin x (-sin x ) /cos x)]

Therefore  f '(x) = (cos x )^{sin x} [(log cos x) . cos x - sin x .tan x]

Question : How to solve this f(x) = x ^{sin x}  

First Assume base x as Log x as 1st function and power function as 2nd function, then apply Product rule of differentiation,and place f(x) in front of the result so obtained.
d/dx { log x . sin x }= ( log x) d/dx ( sin x ) + ( sin x ) d/dx ( log x)

= log x . cos x + sin x . (1/x)                            
                     
Now put f(x) in front of this result and that will be derivative of the f(x).Hence f ' (x) = x ^{sin x} { log x . cos x + sin x . (1/x) }

Question : Differentiate w.r.t. 'x'

 f(x) = cos x ^{sin x} + (sin x) ^x 

Let  f(x) = g(x) +h(x)
Then   f '(x) =g'(x) +h'(x)


Just place cos x ^{sin x}  in front of derivative of {(log cos x) . (sin x) } + place  (sin x) ^x  in front of derivative of { ( log sin x) . ( x) },

So Answer will be g ' (x) = cos x ^{sin x}   { log cos x . cos x +  sin x . (- sin x ) /cos x } 

and h'(x) =  (sin x) ^x { log sin x . (1/x) + x . cos x }

Similarly derivative of h(x) = (sin x) ^x    in one step can also be written as
h '(x) = (sin x) ^x [ log sin x × 1+ x . cos x/sin x ]

Question : Differentiate f (x) = e ^{sin x} 

 then using short cut method , 

f '(x) = e ^{sin x}  [ log e . d/dx{sin x} + sin x d/dx{log  e}]

f '(x) = e ^{sin x}  [ log e × cos x ] ,

Because derivative of log e is zero 

Question : Differentiate f (x) = a ^{sin x} 

If f (x) = a ^{sin x} 

then using short cut method , 

f '(x) =a ^{sin x} [ log a . d/dx{sin x} + sin x d/dx{log  a}]

f '(x) = a ^{sin x} [ log a . cos x ] ,

Because derivative of log a is zero 

Question : Differentiate f (x) = x ^{sin x + cos x} 

If f (x) = x ^{sin x + cos x} 

then using short cut method , 

f '(x) = x ^{sin x + cos x} [ log x . d/dx{sin x + cos x } + {sin x + cos x }d/dx{log x}]

f '(x) = x ^{sin x + cos x} [ log x  . {cos x - sin x} + {sin x+cos x }.{1/x} ] 

One more shortcut for differentiation you can use

            

Conclusion

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