Using First Principal Prove the derivative ?

#root(3)sinx# = #1/3cosx(sinx)^(-2/3)#

1 Answer
Jun 20, 2018

Please see below.

Explanation:

preparation of mind:

Let, #f(x)=(sinx)^(1/3)=>f(t)=(sint)^(1/3)#

#a^3-b^3=(a-b)(a^2+ab+b^2)#

Putting , #(sint)^(1/3)=a and (sinx)^(1/3)=b=>sint=a^3 and sinx=b^3#

#:.sint-sinx=#

#((sint)^(1/3)-(sinx)^(1/3))color(red)([(sint)^(2/3)+(sint)^(1/3) (sinx)^(1/3)+(sinx)^(2/3)]#

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ANSWER:

Differentiating from first principal :

#f'(x)=lim_(t tox)(f(t)-f(x))/(t-x)#

#=lim_(t tox)((sint)^(1/3)-(sinx)^(1/3))/(t-x)#

Multiplying numerator and denominator ,both by:

#color(red) ([(sint)^(2/3)+(sint)^(1/3)(sinx)^(1/3)+(sinx)^(2/3)]# ,we get

#lim_(t tox)((sint)^(1/3)-(sinx)^(1/3))/(t-x)color(red)([(sint)^(2/3)+(sint)^(1/3)(sinx)^(1/3)+(sinx)^(2/3)]/color(red)([(sint)^(2/3)+(sint)^(1/3)(sinx)^(1/3)+(sinx)^(2/3)]#

#f'(x)#=#lim_(t tox)(sint-sinx)/((t-x)(color(red)([(sint)^(2/3)+(sint)^(1/3)(sinx)^(1/3)+(sinx)^(2/3)]))#

#=lim_(t tox)(2cos((t+x)/2)sin((t-x)/2))/(2((t-x)/2))*{1/((sinx)^(2/3)+(sinx)^(1/3+1/3)+(sinx)^(2/3))}#

#=(cos((x+x)/2))/({(sinx)^(2/3)+(sinx)^(2/3)+(sinx)^(2/3)})xxlim_((t-x)/2 to0)(sin((t-x)/2)/((t-x)/2))#

#=cosx/(3(sinx)^(2/3))xx(1)#

#=1/3cosx(sinx)^(-2/3)#