Differential Calculus> find dy/dx when y= e^(ax) . sec x . log x...

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

muthhu swami , 10 Years ago

Grade 10

1 Answers

Arun Kumar

Last Activity: 10 Years ago

Hello Student,

$\\ \\ \frac{d}{dx}\left(e^{ax}\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\right) \\ =\frac{d}{dx}\left(e^{ax}\right)\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\frac{d}{dx}\left(\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\right)e^{ax} \\ \frac{d}{dx}\left(\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\right) \\ =\frac{d}{dx}\left(\sec \left(x\right)\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\frac{d}{dx}\left(\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\right)\sec \left(x\right) \\ =\frac{\tan \left(x\right)}{\cos \left(x\right)}\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\frac{1}{2x\ln \left(10\right)+\frac{\ln \left(10\right)}{x-1}+\ln \left(10\right)}\sec \left(x\right)$

$\\ =e^{ax}a\sec \left(x\right)\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\left(\frac{\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\tan \left(x\right)}{\cos \left(x\right)}+\frac{\sec \left(x\right)}{2x\ln \left(10\right)+\frac{\ln \left(10\right)}{x-1}+\ln \left(10\right)}\right)e^{ax} \\ =e^{ax}\left(\sec \left(x\right)\left(a\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)+\frac{1}{2x\ln \left(10\right)+\frac{\ln \left(10\right)}{x-1}+\ln \left(10\right)}\right)+\frac{\log _{10}\left(\frac{x}{\sqrt{1-2x}}\right)\tan \left(x\right)}{\cos \left(x\right)}\right) \\$