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Derivative of cube root of tanx by first principle |
| Answer» Ans.\xa0\\(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}\\) | |