Differentiate ³√tanx from first principle ?

2 Answers
maganbhai P.
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) )#

Ratnaker Mehta
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!

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