Calculus- Derivatives

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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

it very easy  just wait i will send you the answer

do you want it on the picture

or

just here

??

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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