Differentiate ³√tanx from first principle ?

2 Answers
Jul 30, 2018

f'(x)=sec^2x/(3(tanx)^(2/3) )

Explanation:

We know that ,

color(red)((1)(a-b)(a^2+ab+b^2)=a^3-b^3

color(blue)((2)tan(A-B)=(tanA-tanB)/(1+tanAtanB)

color(brown)((3)lim_(theta to0)(tantheta)/(theta)=1

Let , f(x)=root(3)tanx=(tanx)^(1/3)=>f(t)=(tant)^(1/3)

Using First Principle :

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

=lim_(t tox)((tant)^(1/3)-(tanx)^(1/3))/(t-x)

Before applying color(red)((1) Multiply numerator and denominator by

[(tant)^(2/3) +(tant)^(1/3)(tanx)^(1/3)+(tanx)^(2/3)]

f'(x)=lim_(t tox){color(red)(((tant)^(1/3)-(tanx)^(1/3)))/(t-x) [color(red)((tant)^(2/3) +(tant)^(1/3)(tanx)^(1/3)+(tanx)^(2/3))}/[(tant)^(2/3) +(tant)^(1/3)(tanx)^(1/3)+(tanx)^(2/3]}

Apply color(red)((1) to numerator

=lim_(t tox)color(red)((tant-tanx))/(t-x)1/[(tant)^(2/3) +(tant)^(1/3)(tanx)^(1/3)+(tanx)^(2/3)

=1/[(tanx)^(2/3) +(tanx)^(1/3)(tanx)^(1/3)+(tanx)^(2/3))lim_(t tox)(tant-tanx)/(t-x)
=1/(3(tanx)^(2/3)) xxlim_(t tox)(tant-tanx)/(t-x)

Before applying color(blue)((2) Multiply numerator and denominator by

(1+tant tanx)

f'(x)=1/(3(tanx)^(2/3)) lim_(t tox)color(blue)((tant-tanx))/(t-x)*(1+tant tanx)/color(blue)((1+tant tanx))

:.f'(x)=1/(3(tanx)^(2/3)) lim_(t tox)(color(blue)((tant-tanx)/(1+tant tanx)))/(t-x)*lim_(t tox)(1+tant tanx)

:.f'(x)=1/(3(tanx)^(2/3))*(1+tanxtanx) lim_(t tox)(color(blue)((tant-tanx)/(1+tant tanx)))/(t-x)

Using color(blue)((2) we get

f'(x)=1/(3(tanx)^(2/3))(1+tan^2x) color(brown)(_((t-x)to0) [tan(t-x)/(t-x)])

:.f'(x)=sec^2x/(3(tanx)^(2/3) ) xxcolor(brown)((1))...tocolor(brown)(Apply(3)

=>f'(x)=sec^2x/(3(tanx)^(2/3) )

Jul 30, 2018

sec^2x/(3tan^(2/3)x).

Explanation:

We know that, f'(x)=lim_(t to x){f(t)-f(x)}/(t-x).

So, if f(x)=root(3)tanx=tan^(1/3)x, then,

f'(x)=lim_(t to x){tan^(1/3)t-tan^(1/3)x)/(t-x),

=lim(tan^(1/3)t-tan^(1/3)x)/(tant-tanx)*(tant-tanx)/(t-x),

i.e., f'(x)=L_1*L_2"............................."(ast)," say, where, "

L_1=lim_(t to x)(tan^(1/3)t-tan^(1/3)x)/(tant-tanx).

Here, let tan^(1/3)t=T and tan^(1/3)x=X.

:." As "t to x, T to X; and, tant=T^3, tanx=X^3.

:. L_1=lim_(T to X)(T-X)/(T^3-X^3),

=lim(T-X)/{(T-X)(T^2+TX+X^2)},

=lim_(T to X) 1/(T^2+TX+X^2),

=1/(X^2+X*X+X^2).

rArr L_1=1/(3X^2)=1/(3tan^(2/3)x).................(ast^1).

L_2=lim_(t to x )(tant-tanx)/(t-x),

=lim(sint/cost-sinx/cosx)/(t-x),

=lim(sintcosx-costsinx)/{(cost)(cosx)(t-x)},

=lim_(t to x){(sin(t-x))/(t-x)}sect*secx,

=1*secx*secx.

rArr L_2=sec^2x....................................(ast^2).

Combining (ast), (ast^1) and (ast^2), we have,

[root(3)tanx]'=sec^2x/(3tan^(2/3)x), as Respected Maganbhai P.

has derived!