Using First Principal Prove the derivative ?

root(3)sinx3sinx = 1/3cosx(sinx)^(-2/3)13cosx(sinx)23

1 Answer
Jun 20, 2018

Please see below.

Explanation:

preparation of mind:

Let, f(x)=(sinx)^(1/3)=>f(t)=(sint)^(1/3)f(x)=(sinx)13f(t)=(sint)13

a^3-b^3=(a-b)(a^2+ab+b^2)a3b3=(ab)(a2+ab+b2)

Putting , (sint)^(1/3)=a and (sinx)^(1/3)=b=>sint=a^3 and sinx=b^3(sint)13=aand(sinx)13=bsint=a3andsinx=b3

:.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)