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Question Number 52006 by peter frank last updated on 01/Jan/19

$$Differentiate{\sin }^{-1}\left[\frac{\ln x}{\cos x}\right]$$$$withrespectto\tan x^2$$

Commented by MJS last updated on 02/Jan/19

$$\mathrm{I}\mathrm{get}\left(\mathrm{with}\mathrm{no}\mathrm{respect}\mathrm{to}\mathrm{anything}\right)$$$$\frac{d}{dx}\left[\arcsin \frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{f^{\prime }g-{fg}^{\prime }}{g\sqrt{g^2-f^2}}=$$$$=\frac{\cos x+x\ln x\sin x}{x\cos x\sqrt{{\left(\cos x\right)}^2-{\left(\ln x\right)}^2}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19

$$u={sin}^{-1}\left(\frac{lnx}{cosx}\right)$$$$v={tanx}^2$$$$\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}$$$$u={sin}^{-1}\left(\frac{lnx}{cosx}\right)$$$$\frac{du}{dx}=\frac{1}{\sqrt{1-{\left(\frac{lnx}{cosx}\right)}^2}}\times \frac{cosx\left(\frac{1}{x}\right)-lnx(-sinx)}{{\left(cosx\right)}^2}$$$$=\frac{du}{dx}=\frac{1}{\sqrt{1-{\left(\frac{lnx}{cosx}\right)}^2}}\times \frac{\frac{cosx}{x}+lnx\left(sinx\right)}{{\left(cosx\right)}^2}$$$$v={tanx}^2$$$$\frac{dv}{dx}={sec}^2\left(x^2\right)\times 2x$$$$\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\left[\frac{1}{\sqrt{1-{\left(\frac{lnx}{cosx}\right)}^2}}\times \frac{\frac{cosx}{x}+sinx\times lnx}{{\left(cosx\right)}^2}\right]\times \frac{1}{2{xsec}^2\left(x^2\right)}$$

Commented by peter frank last updated on 02/Jan/19

$$thankyousir.$$

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