Question

find dy/DX of y = e^x.sec x - x^5/3.log x​

Answer 1

Topic :-

Differentiation

To Differentiate :-

$y=e^x\cdot\sec x-\dfrac{x^5}{3\log x}$

Solution :-

$y=e^x\cdot\sec x-\dfrac{x^5}{3\log x}$

$\dfrac{dy}{dx}=\dfrac{d\left(e^x\cdot\sec x-\dfrac{x^5}{3\log x}\right)}{dx}$

$\dfrac{dy}{dx}=\dfrac{d(e^x\cdot\sec x)}{dx}-\dfrac{d}{dx}\left(\dfrac{x^5}{3\log x}\right)$

$\left(\because \dfrac{d(f\pm g)}{dx}=\dfrac{df}{dx}\pm \dfrac{dg}{dx}\right)$

$\dfrac{dy}{dx}=\left( e^x\cdot\dfrac{d(\sec x)}{dx}+\sec x\cdot\dfrac{d(e^x)}{dx}\right)-\dfrac{d}{dx}\left(\dfrac{x^5}{3\log x}\right)$

$\left(\because \dfrac{d(fg)}{dx}=f\cdot\dfrac{dg}{dx}+g\cdot\dfrac{df}{dx}\right)$

$\dfrac{dy}{dx}=\left( e^x\cdot\sec x\cdot \tan x+\sec x\cdot\dfrac{d(e^x)}{dx}\right)-\dfrac{d}{dx}\left(\dfrac{x^5}{3\log x}\right)$

$\left( \because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x\right)$

$\dfrac{dy}{dx}=\left( e^x\cdot\sec x\cdot \tan x+\sec x\cdot e^x\right)-\dfrac{d}{dx}\left(\dfrac{x^5}{3\log x}\right)$

$\left( \because \dfrac{d(e^x)}{dx}=e^x\right)$

$\dfrac{dy}{dx}=\left( e^x\cdot\sec x\cdot \tan x+\sec x\cdot e^x\right)-\dfrac{1}{3}\cdot\dfrac{d}{dx}\left(\dfrac{x^5}{\log x}\right)$

$\left( \because \dfrac{d(k\cdot f(x))}{dx}=k\cdot\dfrac{d(f(x))}{dx},where\:k\:is\:constant.\right)$

$\dfrac{dy}{dx}=e^x\cdot\sec x\left(\tan x+1\right)-\dfrac{1}{3}\left( \dfrac{\log x\cdot\dfrac{d(x^5)}{dx}-x^5\cdot \dfrac{d(\log x)}{dx}}{(\log x)^2}\right)$

$\left(\because \dfrac{d}{dx}\left(\dfrac{f}{g}\right)=\dfrac{g\cdot \dfrac{df}{dx}-f\cdot \dfrac{dg}{dx}}{g^2}\right)$

$\dfrac{dy}{dx}=e^x\cdot\sec x\left(\tan x+1\right)-\dfrac{1}{3}\left( \dfrac{5x^4\log x -x^5\cdot \dfrac{d(\log x)}{dx}}{\log^2 x}\right)$

$\left(\because \dfrac{d(x^n)}{dx}=nx^{n-1}\right)$

$\dfrac{dy}{dx}=e^x\cdot\sec x\left(\tan x+1\right)-\dfrac{1}{3}\left( \dfrac{5x^4\log x -x^5\cdot \dfrac{1}{x}}{\log^2 x}\right)$

$\left(\because \dfrac{d(\log x)}{dx}=\dfrac{1}{x}\right)$

$\dfrac{dy}{dx}=e^x\cdot\sec x\left(\tan x+1\right)-\left( \dfrac{5x^4\log x -x^4}{3\log^2 x}\right)$

Answer :-

$\underline{\boxed{\dfrac{dy}{dx}=e^x\cdot\sec x\left(1+\tan x\right)-\left( \dfrac{5x^4\log x -x^4}{3\log^2 x}\right)}}$