Question Number 52006 by peter frank last updated on 01/Jan/19
![Differentiate sin^(−1) [((ln x)/(cos x))] with respect to tan x^2](https://www.tinkutara.com/question/Q52006.png)
$${Differentiate}\:\mathrm{sin}^{−\mathrm{1}} \left[\frac{\mathrm{ln}\:{x}}{\mathrm{cos}\:{x}}\right] \\ $$$${with}\:{respect}\:{to}\:\mathrm{tan}\:{x}^{\mathrm{2}} \\ $$
Commented by MJS last updated on 02/Jan/19
![I get (with no respect to anything) (d/dx)[arcsin ((f(x))/(g(x)))]=((f′g−fg′)/(g(√(g^2 −f^2 ))))= =((cos x +xln x sin x)/(xcos x (√((cos x)^2 −(ln x)^2 ))))](https://www.tinkutara.com/question/Q52026.png)
$$\mathrm{I}\:\mathrm{get}\:\left(\mathrm{with}\:\mathrm{no}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{anything}\right) \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{arcsin}\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\right]=\frac{{f}'{g}−{fg}'}{{g}\sqrt{{g}^{\mathrm{2}} −{f}^{\mathrm{2}} }}= \\ $$$$=\frac{\mathrm{cos}\:{x}\:+{x}\mathrm{ln}\:{x}\:\mathrm{sin}\:{x}}{{x}\mathrm{cos}\:{x}\:\sqrt{\left(\mathrm{cos}\:{x}\right)^{\mathrm{2}} −\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \:}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19
![u=sin^(−1) (((lnx)/(cosx))) v=tanx^2 (du/dv)=((du/dx)/(dv/dx)) u=sin^(−1) (((lnx)/(cosx))) (du/dx)=(1/( (√(1−(((lnx)/(cosx)))^2 ))))×((cosx((1/x))−lnx(−sinx))/((cosx)^2 )) =(du/dx)=(1/( (√(1−(((lnx)/(cosx)))^2 ))))×((((cosx)/x)+lnx(sinx))/((cosx)^2 )) v=tanx^2 (dv/dx)=sec^2 (x^2 )×2x (du/dv)=((du/dx)/(dv/dx))=[(1/( (√(1−(((lnx)/(cosx)))^2 ))))×((((cosx)/x)+sinx×lnx)/((cosx)^2 ))]×(1/(2xsec^2 (x^2 )))](https://www.tinkutara.com/question/Q52010.png)
$${u}={sin}^{−\mathrm{1}} \left(\frac{{lnx}}{{cosx}}\right) \\ $$$${v}={tanx}^{\mathrm{2}} \\ $$$$\frac{{du}}{{dv}}=\frac{\frac{{du}}{{dx}}}{\frac{{dv}}{{dx}}} \\ $$$${u}={sin}^{−\mathrm{1}} \left(\frac{{lnx}}{{cosx}}\right) \\ $$$$\frac{{du}}{{dx}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\frac{{lnx}}{{cosx}}\right)^{\mathrm{2}} }}×\frac{{cosx}\left(\frac{\mathrm{1}}{{x}}\right)−{lnx}\left(−{sinx}\right)}{\left({cosx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{du}}{{dx}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\frac{{lnx}}{{cosx}}\right)^{\mathrm{2}} }}×\frac{\frac{{cosx}}{{x}}+{lnx}\left({sinx}\right)}{\left({cosx}\right)^{\mathrm{2}} } \\ $$$${v}={tanx}^{\mathrm{2}} \\ $$$$\frac{{dv}}{{dx}}={sec}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)×\mathrm{2}{x} \\ $$$$\frac{{du}}{{dv}}=\frac{\frac{{du}}{{dx}}}{\frac{{dv}}{{dx}}}=\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\frac{{lnx}}{{cosx}}\right)^{\mathrm{2}} }}×\frac{\frac{{cosx}}{{x}}+{sinx}×{lnx}}{\left({cosx}\right)^{\mathrm{2}} }\right]×\frac{\mathrm{1}}{\mathrm{2}{xsec}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)} \\ $$
Commented by peter frank last updated on 02/Jan/19
$${thank}\:{you}\:{sir}. \\ $$