Question

%7Bx%7D%20%20%2B%20%20%7B%20log_%7Be%7Dx%20%7D%20%20%5C%3A%20%20%5C%3A%20%20find%20%5C%3A%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20" id="TexFormula1" title=" \rm \: If \: y = {e}^{x} - \frac{1}{x} + { log_{e}x } \: \: find \: \frac{dy}{dx} " alt=" \rm \: If \: y = {e}^{x} - \frac{1}{x} + { log_{e}x } \: \: find \: \frac{dy}{dx} " align="absmiddle" class="latex-formula">
Thanks!!​

Answer 1

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

• ${\boxed{\bf{\dfrac{d}{dx}\{f(x)\pm g(x)\pm h(x)\pm \cdots =\dfrac{d}{dx}\{f(x)\}\pm \dfrac{d}{dx}\{g(x)\}\pm \dfrac{d}{dx}\{h(x)\}\pm \cdots}}}$

• ${\boxed{\bf{\dfrac{d}{dx}(x^n)=nx^{n-1}}}}$

• ${\boxed{\bf{\dfrac{d}{dx}(log_{e}x)=\dfrac{1}{x}}}}$

• ${\boxed{\bf{\dfrac{d}{dx}(e^x)=e^{x}}}}$

Now consider :

${:\implies \quad \sf y=e^{x}-\dfrac{1}{x}+log_{e}x}$

Differentiating both sides w.r.t.x

${:\implies \quad \sf \dfrac{d}{dx}(y)=\dfrac{d}{dx}\bigg\{e^{x}-\dfrac{1}{x}+log_{e}x\bigg\}}$

${:\implies \quad \sf \dfrac{dy}{dx}=\dfrac{d}{dx}(e^x)-\dfrac{d}{dx}\left(\dfrac{1}{x}\right)+\dfrac{d}{dx}(log_{e}x)}$

${:\implies \quad \sf \dfrac{dy}{dx}=e^{x}-\dfrac{d}{dx}(x^{-1})+\dfrac{1}{x}}$

${:\implies \quad \sf \dfrac{dy}{dx}=e^{x}-(-1)(x)^{-1-1}+\dfrac{1}{x}}$

${:\implies \quad \sf \dfrac{dy}{dx}=e^{x}+(x)^{-2}+\dfrac{1}{x}}$

${:\implies \quad \sf \dfrac{dy}{dx}=e^{x}+\dfrac{1}{x^{2}}+\dfrac{1}{x}}$

Simplifying will yield

${:\implies \quad \bf \therefore \quad \underline{\underline{\dfrac{dy}{dx}=\dfrac{e^{x}x^{2}+x^{2}+x}{x^{2}}}}}$

This is the required answer

Answer 2

Answer:

Given That:

${ \large \longrightarrow{ \rm{ y = {e}^{x} - \frac{1}{x} + { log_{e}x }}}}$

On differentiating partially w.r.t. x, we get:

${\large { \longrightarrow{ \rm{\frac{d}{dx} y = \frac{d}{dx} \left( {e}^{x} - \frac{1}{x} + log_ex \right)}}}}$

${ \large{ \longrightarrow { \rm{\frac{dy}{dx} = \frac{d}{dx} {e}^{x} - \frac{d}{dx} \frac{1}{x} + \frac{d}{dx} log_ex}}}}$

${ \large{ \longrightarrow { \rm{ \frac{dy}{dx} = {e}^{x} - \frac{d}{dx} {x}^{ - 1} + \frac{1}{x} }}}}$

${ \large{ \longrightarrow{ \rm{ \frac{dy}{dx} = {e}^{x} - [ - {x}^{ - 1 - 1} ] + \frac{1}{x} }}}}$

${ \large{ \longrightarrow{ \rm{ \frac{dy}{dx} = {e}^{x} - [ - {x}^{ - 2} ] + \frac{1}{x} }}}}$

Hence:

${ \large \longrightarrow{ \green{ \boxed{{ \rm{ \frac{dy}{dx} = {e}^{x} + \frac{1}{ {x}^{2} } + \frac{1}{x} }}}}}}$

$\:$

Learn More:

${\pink{\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}}}$