Find $left(arctanleft(1+x^2right)right)'$

up vote
1
down vote

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I want to find the derivative of

$$left(arctanleft(1+x^2right)right)'$$

From the derivatives table I see that

$$arctanu=frac 1 1+u^2$$

Therefore it is intuitive for me to replace $1+x^2$ (which is the argument of the function above) with the $u$ of the right side:

$$left(arctanleft(1+x^2right)right)'=frac11+(1+x^2)^2$$

This is not the right solution (the numerator should be equal to $2x$). I understand that $(1+x^2)'=2x$ though I don't know why we have to put $2x$ at the top. Any hints?

edited Aug 23 at 9:10

Bernard

111k635103

asked Aug 23 at 9:04

Cesare

636310

2

Note that $(gcirc f)'(x)=g'(f(x))cdot colorredf'(x).$
â€“Â mfl
Aug 23 at 9:06

Thanks, I didn't study that yet @mfl.
â€“Â Cesare
Aug 23 at 9:07

add a commentÂ |Â

up vote
1
down vote

favorite

I want to find the derivative of

$$left(arctanleft(1+x^2right)right)'$$

From the derivatives table I see that

$$arctanu=frac 1 1+u^2$$

Therefore it is intuitive for me to replace $1+x^2$ (which is the argument of the function above) with the $u$ of the right side:

$$left(arctanleft(1+x^2right)right)'=frac11+(1+x^2)^2$$

This is not the right solution (the numerator should be equal to $2x$). I understand that $(1+x^2)'=2x$ though I don't know why we have to put $2x$ at the top. Any hints?

edited Aug 23 at 9:10

Bernard

111k635103

asked Aug 23 at 9:04

Cesare

636310

2

Note that $(gcirc f)'(x)=g'(f(x))cdot colorredf'(x).$
â€“Â mfl
Aug 23 at 9:06

Thanks, I didn't study that yet @mfl.
â€“Â Cesare
Aug 23 at 9:07

add a commentÂ |Â

up vote
1
down vote

favorite

I want to find the derivative of

$$left(arctanleft(1+x^2right)right)'$$

From the derivatives table I see that

$$arctanu=frac 1 1+u^2$$

Therefore it is intuitive for me to replace $1+x^2$ (which is the argument of the function above) with the $u$ of the right side:

$$left(arctanleft(1+x^2right)right)'=frac11+(1+x^2)^2$$

This is not the right solution (the numerator should be equal to $2x$). I understand that $(1+x^2)'=2x$ though I don't know why we have to put $2x$ at the top. Any hints?

edited Aug 23 at 9:10

Bernard

111k635103

asked Aug 23 at 9:04

Cesare

636310

I want to find the derivative of

$$left(arctanleft(1+x^2right)right)'$$

From the derivatives table I see that

$$arctanu=frac 1 1+u^2$$

Therefore it is intuitive for me to replace $1+x^2$ (which is the argument of the function above) with the $u$ of the right side:

$$left(arctanleft(1+x^2right)right)'=frac11+(1+x^2)^2$$

This is not the right solution (the numerator should be equal to $2x$). I understand that $(1+x^2)'=2x$ though I don't know why we have to put $2x$ at the top. Any hints?

edited Aug 23 at 9:10

Bernard

111k635103

asked Aug 23 at 9:04

Cesare

636310

edited Aug 23 at 9:10

Bernard

111k635103

edited Aug 23 at 9:10

Bernard

111k635103

edited Aug 23 at 9:10

Bernard

111k635103

asked Aug 23 at 9:04

Cesare

636310

asked Aug 23 at 9:04

Cesare

636310

asked Aug 23 at 9:04

Cesare

636310

2

Note that $(gcirc f)'(x)=g'(f(x))cdot colorredf'(x).$
â€“Â mfl
Aug 23 at 9:06

Thanks, I didn't study that yet @mfl.
â€“Â Cesare
Aug 23 at 9:07

add a commentÂ |Â

2

Note that $(gcirc f)'(x)=g'(f(x))cdot colorredf'(x).$
â€“Â mfl
Aug 23 at 9:06

Thanks, I didn't study that yet @mfl.
â€“Â Cesare
Aug 23 at 9:07

Note that $(gcirc f)'(x)=g'(f(x))cdot colorredf'(x).$
â€“Â mfl
Aug 23 at 9:06

Thanks, I didn't study that yet @mfl.
â€“Â Cesare
Aug 23 at 9:07

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
3
down vote

accepted

Hint:

$f(g(x))'=f'(g(x))g'(x)$

answered Aug 23 at 9:06

giannispapav

1,272222

add a commentÂ |Â

up vote
1
down vote

Hint:

use the chain rule of derivatives.

Define $f(x)=arctan(x)$ and $g(x)=1+x^2$. Then, $$arctan(1+x^2)=f(g(x))=(fcirc g)(x)$$

and you can use the fact that $(fcirc g)'(x) = (fcirc g')(x)cdot g'(x)$

answered Aug 23 at 9:06

5xum

82.5k383147

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up vote
1
down vote

$y= arc tan (1+x^2)$

$tan y =1+x^2$

Taking derivative of both sides we get:

$y'(1+tan^2 y)= 2x$

$(arc tan (1+x^2))'=y'=frac2x1+(1+x^2)^2$

answered Aug 23 at 13:33

sirous

896511

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
3
down vote

accepted

Hint:

$f(g(x))'=f'(g(x))g'(x)$

answered Aug 23 at 9:06

giannispapav

1,272222

add a commentÂ |Â

up vote
3
down vote

accepted

Hint:

$f(g(x))'=f'(g(x))g'(x)$

answered Aug 23 at 9:06

giannispapav

1,272222

add a commentÂ |Â

up vote
3
down vote

accepted

Hint:

$f(g(x))'=f'(g(x))g'(x)$

answered Aug 23 at 9:06

giannispapav

1,272222

Hint:

$f(g(x))'=f'(g(x))g'(x)$

answered Aug 23 at 9:06

giannispapav

1,272222

answered Aug 23 at 9:06

giannispapav

1,272222

answered Aug 23 at 9:06

giannispapav

1,272222

answered Aug 23 at 9:06

giannispapav

1,272222

add a commentÂ |Â

up vote
1
down vote

Hint:

use the chain rule of derivatives.

Define $f(x)=arctan(x)$ and $g(x)=1+x^2$. Then, $$arctan(1+x^2)=f(g(x))=(fcirc g)(x)$$

and you can use the fact that $(fcirc g)'(x) = (fcirc g')(x)cdot g'(x)$

answered Aug 23 at 9:06

5xum

82.5k383147

add a commentÂ |Â

up vote
1
down vote

Hint:

use the chain rule of derivatives.

Define $f(x)=arctan(x)$ and $g(x)=1+x^2$. Then, $$arctan(1+x^2)=f(g(x))=(fcirc g)(x)$$

and you can use the fact that $(fcirc g)'(x) = (fcirc g')(x)cdot g'(x)$

answered Aug 23 at 9:06

5xum

82.5k383147

add a commentÂ |Â

up vote
1
down vote

Hint:

use the chain rule of derivatives.

Define $f(x)=arctan(x)$ and $g(x)=1+x^2$. Then, $$arctan(1+x^2)=f(g(x))=(fcirc g)(x)$$

and you can use the fact that $(fcirc g)'(x) = (fcirc g')(x)cdot g'(x)$

answered Aug 23 at 9:06

5xum

82.5k383147

Hint:

use the chain rule of derivatives.

Define $f(x)=arctan(x)$ and $g(x)=1+x^2$. Then, $$arctan(1+x^2)=f(g(x))=(fcirc g)(x)$$

and you can use the fact that $(fcirc g)'(x) = (fcirc g')(x)cdot g'(x)$

answered Aug 23 at 9:06

5xum

82.5k383147

answered Aug 23 at 9:06

5xum

82.5k383147

answered Aug 23 at 9:06

5xum

82.5k383147

answered Aug 23 at 9:06

5xum

82.5k383147

add a commentÂ |Â

up vote
1
down vote

$y= arc tan (1+x^2)$

$tan y =1+x^2$

Taking derivative of both sides we get:

$y'(1+tan^2 y)= 2x$

$(arc tan (1+x^2))'=y'=frac2x1+(1+x^2)^2$

answered Aug 23 at 13:33

sirous

896511

add a commentÂ |Â

up vote
1
down vote

$y= arc tan (1+x^2)$

$tan y =1+x^2$

Taking derivative of both sides we get:

$y'(1+tan^2 y)= 2x$

$(arc tan (1+x^2))'=y'=frac2x1+(1+x^2)^2$

answered Aug 23 at 13:33

sirous

896511

add a commentÂ |Â

up vote
1
down vote

$y= arc tan (1+x^2)$

$tan y =1+x^2$

Taking derivative of both sides we get:

$y'(1+tan^2 y)= 2x$

$(arc tan (1+x^2))'=y'=frac2x1+(1+x^2)^2$

answered Aug 23 at 13:33

sirous

896511

$y= arc tan (1+x^2)$

$tan y =1+x^2$

Taking derivative of both sides we get:

$y'(1+tan^2 y)= 2x$

$(arc tan (1+x^2))'=y'=frac2x1+(1+x^2)^2$

answered Aug 23 at 13:33

sirous

896511

answered Aug 23 at 13:33

sirous

896511

answered Aug 23 at 13:33

sirous

896511

answered Aug 23 at 13:33

sirous

896511

add a commentÂ |Â

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