Question

I am stuck on this.

Let
`y=e^-x+ln(x^2+1)`

Find
`dy/dx`
Find
`d^2y/dx^2`

Answer 1

`\quad \huge \quad \quad \boxed{ \tt \:Answer }`

`\qquad \displaystyle \: \tt \rightarrow \: (dy)/(dx) = - {e}^( - x) + \frac{2x}{ {x}^(2) + 1}`

and

`\qquad \displaystyle \: \tt \rightarrow \: \frac{d {}^(2) {y}^{} }{dx {}^(2) } = {e}^(-1) - \frac{ 2{x}^(2) - 2 }{x {}^(4) + 2 {x}^(2) + 1{}^{} }`

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`\large \tt Solution \: :`

`\qquad \tt \rightarrow \: y = {e}^( - x) + ln( {x}^(2) + 1)`

`\qquad \displaystyle \: \tt \rightarrow \: (dy)/(dx) = (d)/(dx) ( {e}^( - x) ) + (d)/(dx) (ln( {x}^(2) + 1))`

`\qquad \displaystyle \: \tt \rightarrow \: (dy)/(dx) = {e}^( - x) \sdot( - 1) + \frac{1}{ {x}^(2) + 1} \sdot(2x)`

`\qquad \displaystyle \: \tt \rightarrow \: (dy)/(dx) = - {e}^( - x) + \frac{2x}{ {x}^(2) + 1}`

Second derivative :

`\qquad \displaystyle \: \tt \rightarrow \: (d)/(dx) \bigg( (dy)/(dx) \bigg) = (d)/(dx) \bigg (- {e}^( - x) + \frac{2x}{ {x}^(2) + 1} \bigg)`

`\qquad \displaystyle \: \tt \rightarrow \: \frac{d {}^(2) {y}^{} }{dx {}^(2) } = - (d)/(dx) ( {e}^(-1) ) + (d)/(dx) \bigg( \frac{2x}{ {x}^(2) + 1} \bigg )`

`\qquad \displaystyle \: \tt \rightarrow \: \frac{d {}^(2) {y}^{} }{dx {}^(2) } = - {e}^(-1) \sdot( - 1) + \frac{( {x}^(2) + 1) \sdot (d)/(dx)(2x) - 2x \sdot (d)/(dx)( {x}^(2) + 1) }{(x {}^(2) + 1) {}^(2) }`

[ by division rule ]

`\qquad \displaystyle \: \tt \rightarrow \: \frac{d {}^(2) {y}^{} }{dx {}^(2) } = {e}^(-1) + \frac{( {x}^(2) + 1) \sdot (2)- 2x \sdot (2x) }{(x {}^(4) + 1 + 2 {x}^(2) ) {}^{} }`

`\qquad \displaystyle \: \tt \rightarrow \: \frac{d {}^(2) {y}^{} }{dx {}^(2) } = {e}^(-1) + \frac{2{x}^(2) + 2- 4 {x}^(2) }{(x {}^(4) + 1 + 2 {x}^(2) ) {}^{} }`

`\qquad \displaystyle \: \tt \rightarrow \: \frac{d {}^(2) {y}^{} }{dx {}^(2) } = {e}^(-1) + \frac{ - 2{x}^(2) + 2 }{x {}^(4) + 2 {x}^(2) + 1{}^{} }`

`\qquad \displaystyle \: \tt \rightarrow \: \frac{d {}^(2) {y}^{} }{dx {}^(2) } = {e}^(-1) - \frac{ 2{x}^(2) - 2 }{x {}^(4) + 2 {x}^(2) + 1{}^{} }`

Answered by : ❝ AǫᴜᴀWɪᴢ ❞