Calculus- Derivatives

Calculus
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Let f(x)=cosx

Prove f^1(x)=sinx by definition

using: Lim h--0 f(x+h)-f(x)/h

Oct 4th, 2015

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it very easy  just wait i will send you the answer 

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Oct 4th, 2015

Picture would be nice!

Oct 4th, 2015

Is there any way you can fit this question as well?..

Let f(x)=x^n

Prove f^1(x)=nx^n-1  by definition 


Oct 4th, 2015

?

Oct 5th, 2015

i will answer you here sorry my camera not work : 

Lim h--0 f(x+h)-f(x)/h = Lim h--0 cos(x+h)-cos(x)/h 

and we know : cos(a + b) = cos(a) cos(b) – sin(a) sin(b) 

 so : cos(x + h) = cos(x) cos(h) – sin(x) sin(h) 

       Lim h--0 cos(x+h)-cos(x)/h = Lim h--0 (cos(x) cos(h) – sin(x) sin(h) -cos(x))/h 

                                               = Lim h--0 cos(x)(cos(h) – sin(x)sin(h) -1)/h  

                                                = Lim h--0( (cos(x)(cos(h))/h – (sin(x)(sin(h) -1)/h )

                                                = cos(x)Lim h--0 ((cos(h))/h –sin(x) Lim h--0((sin(h) -1)/h 


   and we know :   Lim h--0(cos(h))/h) = 0   and :  Lim h--0((sin(h) -1)/h) = 1 

so :                  Lim h--0 cos(x+h)-cos(x)/h  =    –sin(x)                   

Oct 5th, 2015

i will answer you here sorry my camera not work : 

Lim h--0 f(x+h)-f(x)/h = Lim h--0 cos(x+h)-cos(x)/h 

and we know : cos(a + b) = cos(a) cos(b) – sin(a) sin(b) 

 so : cos(x + h) = cos(x) cos(h) – sin(x) sin(h) 

       Lim h--0 cos(x+h)-cos(x)/h = Lim h--0 (cos(x) cos(h) – sin(x) sin(h) -cos(x))/h 

                                               = Lim h--0 cos(x)(cos(h) – sin(x)sin(h) -1)/h  

                                                = Lim h--0( (cos(x)(cos(h))/h – (sin(x)(sin(h) -1)/h )

                                                = cos(x)Lim h--0 ((cos(h))/h –sin(x) Lim h--0((sin(h) -1)/h 


   and we know :   Lim h--0(cos(h))/h) = 0   and :  Lim h--0((sin(h) -1)/h) = 1 

so :                  Lim h--0 cos(x+h)-cos(x)/h  =    –sin(x)                   

Oct 5th, 2015

for the second the answer is here but on frensh : 

http://www.ilemaths.net/sujet-demontration-de-la-derivee-10-puissance-n-208747.html

Oct 5th, 2015

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