Question

`\rm \: If \: y = {e}^(x) - (1)/(x) + { log_(e)x } \: \: find \: (dy)/(dx)`

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{(d)/(dx)\{f(x)\pm g(x)\pm h(x)\pm \cdots =(d)/(dx)\{f(x)\}\pm (d)/(dx)\{g(x)\}\pm (d)/(dx)\{h(x)\}\pm \cdots}}}`

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

• 
  `{\boxed{\bf{(d)/(dx)(log_(e)x)=(1)/(x)}}}`

• 
  `{\boxed{\bf{(d)/(dx)(e^x)=e^(x)}}}`

Now consider :

`{:\implies \quad \sf y=e^(x)-(1)/(x)+log_(e)x}`

Differentiating both sides w.r.t.x

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

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

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

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

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

`{:\implies \quad \sf (dy)/(dx)=e^(x)+(1)/(x^(2))+(1)/(x)}`

Simplifying will yield

`{:\implies \quad \bf \therefore \quad \underline{\underline{(dy)/(dx)=(e^(x)x^(2)+x^(2)+x)/(x^(2))}}}`

This is the required answer