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)
