Respuesta :
We are given with a function y and need to find dy/dx or the first Derivative of y w.r.t.x , but let's recall
- [tex]{\boxed{\bf{\dfrac{d}{dx}\{f(x)\pm g(x)\pm h(x)\pm \cdots =\dfrac{d}{dx}\{f(x)\}\pm \dfrac{d}{dx}\{g(x)\}\pm \dfrac{d}{dx}\{h(x)\}\pm \cdots}}}[/tex]
- [tex]{\boxed{\bf{\dfrac{d}{dx}(x^n)=nx^{n-1}}}}[/tex]
- [tex]{\boxed{\bf{\dfrac{d}{dx}(log_{e}x)=\dfrac{1}{x}}}}[/tex]
- [tex]{\boxed{\bf{\dfrac{d}{dx}(e^x)=e^{x}}}}[/tex]
Now consider :
[tex]{:\implies \quad \sf y=e^{x}-\dfrac{1}{x}+log_{e}x}[/tex]
Differentiating both sides w.r.t.x
[tex]{:\implies \quad \sf \dfrac{d}{dx}(y)=\dfrac{d}{dx}\bigg\{e^{x}-\dfrac{1}{x}+log_{e}x\bigg\}}[/tex]
[tex]{:\implies \quad \sf \dfrac{dy}{dx}=\dfrac{d}{dx}(e^x)-\dfrac{d}{dx}\left(\dfrac{1}{x}\right)+\dfrac{d}{dx}(log_{e}x)}[/tex]
[tex]{:\implies \quad \sf \dfrac{dy}{dx}=e^{x}-\dfrac{d}{dx}(x^{-1})+\dfrac{1}{x}}[/tex]
[tex]{:\implies \quad \sf \dfrac{dy}{dx}=e^{x}-(-1)(x)^{-1-1}+\dfrac{1}{x}}[/tex]
[tex]{:\implies \quad \sf \dfrac{dy}{dx}=e^{x}+(x)^{-2}+\dfrac{1}{x}}[/tex]
[tex]{:\implies \quad \sf \dfrac{dy}{dx}=e^{x}+\dfrac{1}{x^{2}}+\dfrac{1}{x}}[/tex]
Simplifying will yield
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{\dfrac{dy}{dx}=\dfrac{e^{x}x^{2}+x^{2}+x}{x^{2}}}}}[/tex]
This is the required answer
