# find dy/dx wheny= e^(ax) . sec x . log x/ sqrt(1 – 2x)

Rinkoo Gupta
9 years ago

y= e^(ax) . sec x . log x/ sqrt(1 – 2x)

Taking eaxsecx as first function andlog x/ sqrt(1 – 2x) as second function we use the product rule

dy/dx=eaxsecx.[{√(1-2x).(1/x)-logx.(1/2√(1-2x) )(-2)}/(1-2x)]+

logx/√(1-2x){eaxsecxtanx+secxeax.a}

Now on simplification you get desired result.

Thanks & Regards

Rinkoo Gupta

Arun Kumar IIT Delhi
9 years ago
Hello
$\\ \frac{d}{dx}\left(e^{ax}\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\right) \\ =\frac{d}{dx}\left(e^{ax}\right)\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\frac{d}{dx}\left(\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\right)e^{ax} \\ \frac{d}{dx}\left(\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\right) \\ =\frac{d}{dx}\left(\sec \left(x\right)\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\frac{d}{dx}\left(\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\right)\sec \left(x\right)$
$\\ =\frac{\tan \left(x\right)}{\cos \left(x\right)}\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\frac{1}{2x\ln \left(10\right)+\frac{\ln \left(10\right)}{x-1}+\ln \left(10\right)}\sec \left(x\right) \\ =e^{ax}a\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\left(\frac{\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\tan \left(x\right)}{\cos \left(x\right)}+\frac{\sec \left(x\right)}{2x\ln \left(10\right)+\frac{\ln \left(10\right)}{x-1}+\ln \left(10\right)}\right)e^{ax} \\ =e^{ax}\left(\sec \left(x\right)\left(a\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\frac{1}{2x\ln \left(10\right)+\frac{\ln \left(10\right)}{x-1}+\ln \left(10\right)}\right)+\frac{\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\tan \left(x\right)}{\cos \left(x\right)}\right)$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi