Derivative of cube root of tanx …

Ans. 

$Derivative \space of \space \sqrt [3] {tan x} \space by \space First \space Principle \space is \space given \space by :$

$=> f'(x) = lim_{h \to 0} \space {{ \sqrt [3]{tan(x+h)} - \sqrt [3] {tan x}} \over h}$

$=> lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h)\over cos(x+h)} - \sqrt [3] {sinx\over cosx }} \over h}$

$=> lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h) cos \space x} - \sqrt [3] {sinx \space cos(x+h) }} \over h {\sqrt [3]{cos(x+h) cosx} }}$

$=> lim_{h \to 0} {1\over {\sqrt [3]{cos(x+h) cosx} }} . \space lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h) cos \space x} - \sqrt [3] {sinx \space cos(x+h) }} \over h}$

$=> {1\over {\sqrt [3]{cos(x+0) cosx} }} . \space lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h) cos \space x} - \sqrt [3] {sinx \space cos(x+h) }} \over h}$

$=> {1\over {\sqrt [3]{cos^2x } }}.\space lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h) cos \space x} - \sqrt [3] {sinx\space cos(x+h)}}\over h} \times{ {[{(sin(x+h) cos \space x})^{2\over 3}}+{({sinx \space cos(x+h)})^{2\over3}}+\sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}]\over {[{(sin(x+h) cos \space x})^{2\over 3}}+{({sinx \space cos(x+h)})^{2\over3}}+ \sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}]}$   

$=> {1\over {\sqrt [3]{cos^2x } }}.\space lim_{h \to 0}{ \space {{sin(x+h) cos \space x - sinx\space cos(x+h)}} \over {h [{(sin(x+h) cos \space x})^{2\over 3}+({sinx \space cos(x+h)})^{2\over3}+ \sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}}]}$                       $[a^3-b^3 = (a-b)(a^2+b^2+ab)]$

$=> {1\over {\sqrt [3]{cos^2x } }}.\space lim_{h \to 0}{ \space {{sin(x+h -x )}} \over {h [{(sin(x+h) cos \space x})^{2\over 3}+({sinx \space cos(x+h)})^{2\over3}+ \sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}}]}$             $[sin (a-b) = sina.cosb - cos a. sin b ]$

$=> {1\over {\sqrt [3]{cos^2x } }}.\space lim_{h \to 0}{ \space {{sinh }} \over h} . \space lim_{h \to 0}{ { 1\over [{(sin(x+h) cos \space x})^{2\over 3}+({sinx \space cos(x+h)})^{2\over3}+ \sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}]}}$

$=> {1\over cos^{2\over3}x} . 1. {1\over[(sin(x+0) cos \space x)^{2\over 3}+(sinx \space cos(x+0))^{2\over3}+ \sqrt [3]{sin(x+0)cosx. \space cox(x+0)sinx}]}$

$=> {1\over cos^{2\over3}x} . 1. {1\over[(sinx cos \space x)^{2\over 3}+(sinx \space cosx)^{2\over3}+ \sqrt [3]{sinx.cosx. \space cosx .sinx}]}$

$=> {1\over cos^{2\over3}x} . 1. {1\over[(sinx cos \space x)^{2\over 3}+(sinx \space cosx)^{2\over3}+ ({sinxcosx})^{2\over3}]}$

$=> {1\over cos^{2\over3}x} . {1\over3(sinx cos \space x)^{2\over 3}}$

$=> {1\over 3}{1\over cos^{4\over3}x. sin^{2\over 3}x}$

$=> {1\over 3}{{1\over cos^2x}\over ({ cos^{4\over3}x. sin^{2\over 3}x\over cos^2x})}$

$=> {1\over 3}{{sec^2x}\over ({ sin^{2\over 3}x\over cos^{2\over 3}x})}$

$=> {1\over 3}.{1\over { tan^{2\over 3}x}}.{sec^2x}$