Calculate the derivative of $f(x)=int_x^x^2tan(x+y),dy$ on the open interval $(0,fracpi4)$.

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This appears to be a chain rule question, but I'm struggling with the setup. My attempt is as follows:

Let $F:mathbbR^2rightarrowmathbbR$,

$F(u,v)=int_u^u^2$tan$(v+y)dy$,

$u:mathbbRrightarrowmathbbR^2$,

and $u(x)=(u(x),v(x))=(x,x)$.

Then writing $f(x)=(Fcirc u)(x)$ allows for use of the chain rule:

$D(Fcirc u)(x)=DF(u(x),v(x))Du(x)=left[fracpartial Fpartial umbox fracpartial Fpartial vright]beginbmatrixu'(x)\v'(x)endbmatrix=fracpartial Fpartial u(u(x))u'(x)+fracpartial Fpartial v(u(x))v'(x)$

Then by the Fundamental Theorem of Calculus,

$fracpartial Fpartial u=fracpartialpartial uint_u^u^2tan(v+y)dy=mboxtan(v+u)=mboxtan(x+x)=mboxtan(2x)$

And similarly,

$fracpartial Fpartial v=fracpartialpartial vint_u^u^2mboxtan(v+y)dy=int_u^u^2fracpartialpartial vmboxtan(v+y)dy=int_x^x^2fracpartialpartial xmboxtan(x+y)dy=int_x^x^2mboxsec^2(x+y)dy$

However at this point, I'm relatively convinced I'm making a major mistake, either with my entire substitution or with the way I'm handling the boundaries. Any advice is much appreciated.

real-analysis multivariable-calculus proof-verification

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edited Jul 25 at 21:17

asked Jul 25 at 21:01

Atsina

513113

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up vote
2
down vote

favorite

This appears to be a chain rule question, but I'm struggling with the setup. My attempt is as follows:

Let $F:mathbbR^2rightarrowmathbbR$,

$F(u,v)=int_u^u^2$tan$(v+y)dy$,

$u:mathbbRrightarrowmathbbR^2$,

and $u(x)=(u(x),v(x))=(x,x)$.

Then writing $f(x)=(Fcirc u)(x)$ allows for use of the chain rule:

$D(Fcirc u)(x)=DF(u(x),v(x))Du(x)=left[fracpartial Fpartial umbox fracpartial Fpartial vright]beginbmatrixu'(x)\v'(x)endbmatrix=fracpartial Fpartial u(u(x))u'(x)+fracpartial Fpartial v(u(x))v'(x)$

Then by the Fundamental Theorem of Calculus,

$fracpartial Fpartial u=fracpartialpartial uint_u^u^2tan(v+y)dy=mboxtan(v+u)=mboxtan(x+x)=mboxtan(2x)$

And similarly,

$fracpartial Fpartial v=fracpartialpartial vint_u^u^2mboxtan(v+y)dy=int_u^u^2fracpartialpartial vmboxtan(v+y)dy=int_x^x^2fracpartialpartial xmboxtan(x+y)dy=int_x^x^2mboxsec^2(x+y)dy$

However at this point, I'm relatively convinced I'm making a major mistake, either with my entire substitution or with the way I'm handling the boundaries. Any advice is much appreciated.

real-analysis multivariable-calculus proof-verification

share|cite|improve this question

edited Jul 25 at 21:17

asked Jul 25 at 21:01

Atsina

513113

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

This appears to be a chain rule question, but I'm struggling with the setup. My attempt is as follows:

Let $F:mathbbR^2rightarrowmathbbR$,

$F(u,v)=int_u^u^2$tan$(v+y)dy$,

$u:mathbbRrightarrowmathbbR^2$,

and $u(x)=(u(x),v(x))=(x,x)$.

Then writing $f(x)=(Fcirc u)(x)$ allows for use of the chain rule:

$D(Fcirc u)(x)=DF(u(x),v(x))Du(x)=left[fracpartial Fpartial umbox fracpartial Fpartial vright]beginbmatrixu'(x)\v'(x)endbmatrix=fracpartial Fpartial u(u(x))u'(x)+fracpartial Fpartial v(u(x))v'(x)$

Then by the Fundamental Theorem of Calculus,

$fracpartial Fpartial u=fracpartialpartial uint_u^u^2tan(v+y)dy=mboxtan(v+u)=mboxtan(x+x)=mboxtan(2x)$

And similarly,

$fracpartial Fpartial v=fracpartialpartial vint_u^u^2mboxtan(v+y)dy=int_u^u^2fracpartialpartial vmboxtan(v+y)dy=int_x^x^2fracpartialpartial xmboxtan(x+y)dy=int_x^x^2mboxsec^2(x+y)dy$

However at this point, I'm relatively convinced I'm making a major mistake, either with my entire substitution or with the way I'm handling the boundaries. Any advice is much appreciated.

real-analysis multivariable-calculus proof-verification

share|cite|improve this question

edited Jul 25 at 21:17

asked Jul 25 at 21:01

Atsina

513113

This appears to be a chain rule question, but I'm struggling with the setup. My attempt is as follows:

Let $F:mathbbR^2rightarrowmathbbR$,

$F(u,v)=int_u^u^2$tan$(v+y)dy$,

$u:mathbbRrightarrowmathbbR^2$,

and $u(x)=(u(x),v(x))=(x,x)$.

Then writing $f(x)=(Fcirc u)(x)$ allows for use of the chain rule:

$D(Fcirc u)(x)=DF(u(x),v(x))Du(x)=left[fracpartial Fpartial umbox fracpartial Fpartial vright]beginbmatrixu'(x)\v'(x)endbmatrix=fracpartial Fpartial u(u(x))u'(x)+fracpartial Fpartial v(u(x))v'(x)$

Then by the Fundamental Theorem of Calculus,

$fracpartial Fpartial u=fracpartialpartial uint_u^u^2tan(v+y)dy=mboxtan(v+u)=mboxtan(x+x)=mboxtan(2x)$

And similarly,

$fracpartial Fpartial v=fracpartialpartial vint_u^u^2mboxtan(v+y)dy=int_u^u^2fracpartialpartial vmboxtan(v+y)dy=int_x^x^2fracpartialpartial xmboxtan(x+y)dy=int_x^x^2mboxsec^2(x+y)dy$

However at this point, I'm relatively convinced I'm making a major mistake, either with my entire substitution or with the way I'm handling the boundaries. Any advice is much appreciated.

real-analysis multivariable-calculus proof-verification

share|cite|improve this question

edited Jul 25 at 21:17

asked Jul 25 at 21:01

Atsina

513113

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share|cite|improve this question

edited Jul 25 at 21:17

edited Jul 25 at 21:17

edited Jul 25 at 21:17

asked Jul 25 at 21:01

Atsina

513113

asked Jul 25 at 21:01

Atsina

513113

asked Jul 25 at 21:01

Atsina

513113

513113

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add a comment | 

4 Answers
4

active

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up vote
4
down vote



accepted

Using Leibniz integral rule:

beginalign
f'(x) &= 2xtan(x+x^2) - tan(2x) + int_x^x^2 fracdycos^2(x+y)\
&= 2xtan(x+x^2) - tan(2x)+ tan(x+y)Big|_x^x^2\
&= (2x+1)tan(x+x^2) - 2tan(2x)
endalign

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answered Jul 25 at 21:26

mechanodroid

22.2k52041

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up vote
1
down vote

Let me try:
beginalign
f'(x)
&=partial_x intlimits_x^x^2tan(x+y),dy \
&= tan(x+ x^2) (x^2)' - tan(x + x)(x)' +
intlimits_x^x^2partial_x tan(x+y),dy \
&= 2x tan(x+ x^2) - tan(2x) +
intlimits_x^x^2(1+ tan^2(x+y)) dy \
&= 2x tan(x+ x^2) - tan(2x) +
left[tan(x + y) right]_y=x^y=x^2 \
&= 2x tan(x+x^2) - tan(2x) + tan(x+x^2) - tan(2x) \
&= (2x+1) tan(x+x^2) - 2tan(2x)
endalign

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edited Jul 25 at 21:31

answered Jul 25 at 21:23

mvw

30.2k22250

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up vote
1
down vote

Based upon your correction, I have revised my answer. $f(x)=int_x^x^2tan(x+y)dy=ln(cos(x+x^2))-ln(cos(2x))$. Therefore $f'(x)=(2x+1)tan(x+x^2)-2tan(2x)$.

share|cite|improve this answer

edited Jul 26 at 0:50

answered Jul 25 at 21:21

herb steinberg

93529

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The question is phrased in terms of $x$ and $y$, not $x$, $v$, and $y$. Not sure why @Holo edited my original post to appear that way, but I changed it back
  – Atsina
  Jul 25 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I believe you've flipped a negative during the integration
  – Atsina
  Jul 25 at 21:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question can be generalized. Let $f(x)=int_a(x)^b(x)h(c(x)+y)dy$, where $a(x),b(x),c(x),h(x)$ are all known. Then $f(x)=H(c(x)+b(x))-H(c(x)+a(x))$ where $H(x)$ may not be explicitly representable, but $H'(x)=h(x)$. Then $f'(x)=(c'(x)+b'(x))h(c(x)+b(x))-(c'(x)+a'(x))(h(c(x)+a(x))$. The chain rule is used at the end.
  – herb steinberg
  Jul 26 at 16:19
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Hint:

Let $$F(x,t)= int_0^x tan(t+y)dy$$

Then we have $$f(x)= F(x,x^2)-F(x,x)$$

NOTE THAT $partial_x F(x,t) = tan(t+x)$ and $partial_t F(x,t) = tan(x+t)- tan(t)$

can you conclude?

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edited Jul 26 at 2:56

answered Jul 26 at 2:46

Guy Fsone

16.8k42671

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Using Leibniz integral rule:

beginalign
f'(x) &= 2xtan(x+x^2) - tan(2x) + int_x^x^2 fracdycos^2(x+y)\
&= 2xtan(x+x^2) - tan(2x)+ tan(x+y)Big|_x^x^2\
&= (2x+1)tan(x+x^2) - 2tan(2x)
endalign

share|cite|improve this answer

answered Jul 25 at 21:26

mechanodroid

22.2k52041

add a comment | 

up vote
4
down vote



accepted

Using Leibniz integral rule:

beginalign
f'(x) &= 2xtan(x+x^2) - tan(2x) + int_x^x^2 fracdycos^2(x+y)\
&= 2xtan(x+x^2) - tan(2x)+ tan(x+y)Big|_x^x^2\
&= (2x+1)tan(x+x^2) - 2tan(2x)
endalign

share|cite|improve this answer

answered Jul 25 at 21:26

mechanodroid

22.2k52041

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Using Leibniz integral rule:

beginalign
f'(x) &= 2xtan(x+x^2) - tan(2x) + int_x^x^2 fracdycos^2(x+y)\
&= 2xtan(x+x^2) - tan(2x)+ tan(x+y)Big|_x^x^2\
&= (2x+1)tan(x+x^2) - 2tan(2x)
endalign

share|cite|improve this answer

answered Jul 25 at 21:26

mechanodroid

22.2k52041

Using Leibniz integral rule:

beginalign
f'(x) &= 2xtan(x+x^2) - tan(2x) + int_x^x^2 fracdycos^2(x+y)\
&= 2xtan(x+x^2) - tan(2x)+ tan(x+y)Big|_x^x^2\
&= (2x+1)tan(x+x^2) - 2tan(2x)
endalign

share|cite|improve this answer

answered Jul 25 at 21:26

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 21:26

mechanodroid

22.2k52041

answered Jul 25 at 21:26

mechanodroid

22.2k52041

answered Jul 25 at 21:26

mechanodroid

22.2k52041

22.2k52041

add a comment | 

add a comment | 

up vote
1
down vote

Let me try:
beginalign
f'(x)
&=partial_x intlimits_x^x^2tan(x+y),dy \
&= tan(x+ x^2) (x^2)' - tan(x + x)(x)' +
intlimits_x^x^2partial_x tan(x+y),dy \
&= 2x tan(x+ x^2) - tan(2x) +
intlimits_x^x^2(1+ tan^2(x+y)) dy \
&= 2x tan(x+ x^2) - tan(2x) +
left[tan(x + y) right]_y=x^y=x^2 \
&= 2x tan(x+x^2) - tan(2x) + tan(x+x^2) - tan(2x) \
&= (2x+1) tan(x+x^2) - 2tan(2x)
endalign

share|cite|improve this answer

edited Jul 25 at 21:31

answered Jul 25 at 21:23

mvw

30.2k22250

add a comment | 

up vote
1
down vote

Let me try:
beginalign
f'(x)
&=partial_x intlimits_x^x^2tan(x+y),dy \
&= tan(x+ x^2) (x^2)' - tan(x + x)(x)' +
intlimits_x^x^2partial_x tan(x+y),dy \
&= 2x tan(x+ x^2) - tan(2x) +
intlimits_x^x^2(1+ tan^2(x+y)) dy \
&= 2x tan(x+ x^2) - tan(2x) +
left[tan(x + y) right]_y=x^y=x^2 \
&= 2x tan(x+x^2) - tan(2x) + tan(x+x^2) - tan(2x) \
&= (2x+1) tan(x+x^2) - 2tan(2x)
endalign

share|cite|improve this answer

edited Jul 25 at 21:31

answered Jul 25 at 21:23

mvw

30.2k22250

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let me try:
beginalign
f'(x)
&=partial_x intlimits_x^x^2tan(x+y),dy \
&= tan(x+ x^2) (x^2)' - tan(x + x)(x)' +
intlimits_x^x^2partial_x tan(x+y),dy \
&= 2x tan(x+ x^2) - tan(2x) +
intlimits_x^x^2(1+ tan^2(x+y)) dy \
&= 2x tan(x+ x^2) - tan(2x) +
left[tan(x + y) right]_y=x^y=x^2 \
&= 2x tan(x+x^2) - tan(2x) + tan(x+x^2) - tan(2x) \
&= (2x+1) tan(x+x^2) - 2tan(2x)
endalign

share|cite|improve this answer

edited Jul 25 at 21:31

answered Jul 25 at 21:23

mvw

30.2k22250

Let me try:
beginalign
f'(x)
&=partial_x intlimits_x^x^2tan(x+y),dy \
&= tan(x+ x^2) (x^2)' - tan(x + x)(x)' +
intlimits_x^x^2partial_x tan(x+y),dy \
&= 2x tan(x+ x^2) - tan(2x) +
intlimits_x^x^2(1+ tan^2(x+y)) dy \
&= 2x tan(x+ x^2) - tan(2x) +
left[tan(x + y) right]_y=x^y=x^2 \
&= 2x tan(x+x^2) - tan(2x) + tan(x+x^2) - tan(2x) \
&= (2x+1) tan(x+x^2) - 2tan(2x)
endalign

share|cite|improve this answer

edited Jul 25 at 21:31

answered Jul 25 at 21:23

mvw

30.2k22250

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 21:31

edited Jul 25 at 21:31

edited Jul 25 at 21:31

answered Jul 25 at 21:23

mvw

30.2k22250

answered Jul 25 at 21:23

mvw

30.2k22250

answered Jul 25 at 21:23

mvw

30.2k22250

30.2k22250

add a comment | 

add a comment | 

up vote
1
down vote

Based upon your correction, I have revised my answer. $f(x)=int_x^x^2tan(x+y)dy=ln(cos(x+x^2))-ln(cos(2x))$. Therefore $f'(x)=(2x+1)tan(x+x^2)-2tan(2x)$.

share|cite|improve this answer

edited Jul 26 at 0:50

answered Jul 25 at 21:21

herb steinberg

93529

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The question is phrased in terms of $x$ and $y$, not $x$, $v$, and $y$. Not sure why @Holo edited my original post to appear that way, but I changed it back
  – Atsina
  Jul 25 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I believe you've flipped a negative during the integration
  – Atsina
  Jul 25 at 21:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question can be generalized. Let $f(x)=int_a(x)^b(x)h(c(x)+y)dy$, where $a(x),b(x),c(x),h(x)$ are all known. Then $f(x)=H(c(x)+b(x))-H(c(x)+a(x))$ where $H(x)$ may not be explicitly representable, but $H'(x)=h(x)$. Then $f'(x)=(c'(x)+b'(x))h(c(x)+b(x))-(c'(x)+a'(x))(h(c(x)+a(x))$. The chain rule is used at the end.
  – herb steinberg
  Jul 26 at 16:19
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Based upon your correction, I have revised my answer. $f(x)=int_x^x^2tan(x+y)dy=ln(cos(x+x^2))-ln(cos(2x))$. Therefore $f'(x)=(2x+1)tan(x+x^2)-2tan(2x)$.

share|cite|improve this answer

edited Jul 26 at 0:50

answered Jul 25 at 21:21

herb steinberg

93529

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The question is phrased in terms of $x$ and $y$, not $x$, $v$, and $y$. Not sure why @Holo edited my original post to appear that way, but I changed it back
  – Atsina
  Jul 25 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I believe you've flipped a negative during the integration
  – Atsina
  Jul 25 at 21:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question can be generalized. Let $f(x)=int_a(x)^b(x)h(c(x)+y)dy$, where $a(x),b(x),c(x),h(x)$ are all known. Then $f(x)=H(c(x)+b(x))-H(c(x)+a(x))$ where $H(x)$ may not be explicitly representable, but $H'(x)=h(x)$. Then $f'(x)=(c'(x)+b'(x))h(c(x)+b(x))-(c'(x)+a'(x))(h(c(x)+a(x))$. The chain rule is used at the end.
  – herb steinberg
  Jul 26 at 16:19
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Based upon your correction, I have revised my answer. $f(x)=int_x^x^2tan(x+y)dy=ln(cos(x+x^2))-ln(cos(2x))$. Therefore $f'(x)=(2x+1)tan(x+x^2)-2tan(2x)$.

share|cite|improve this answer

edited Jul 26 at 0:50

answered Jul 25 at 21:21

herb steinberg

93529

Based upon your correction, I have revised my answer. $f(x)=int_x^x^2tan(x+y)dy=ln(cos(x+x^2))-ln(cos(2x))$. Therefore $f'(x)=(2x+1)tan(x+x^2)-2tan(2x)$.

share|cite|improve this answer

edited Jul 26 at 0:50

answered Jul 25 at 21:21

herb steinberg

93529

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 0:50

edited Jul 26 at 0:50

edited Jul 26 at 0:50

answered Jul 25 at 21:21

herb steinberg

93529

answered Jul 25 at 21:21

herb steinberg

93529

answered Jul 25 at 21:21

herb steinberg

93529

93529

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The question is phrased in terms of $x$ and $y$, not $x$, $v$, and $y$. Not sure why @Holo edited my original post to appear that way, but I changed it back
  – Atsina
  Jul 25 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I believe you've flipped a negative during the integration
  – Atsina
  Jul 25 at 21:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question can be generalized. Let $f(x)=int_a(x)^b(x)h(c(x)+y)dy$, where $a(x),b(x),c(x),h(x)$ are all known. Then $f(x)=H(c(x)+b(x))-H(c(x)+a(x))$ where $H(x)$ may not be explicitly representable, but $H'(x)=h(x)$. Then $f'(x)=(c'(x)+b'(x))h(c(x)+b(x))-(c'(x)+a'(x))(h(c(x)+a(x))$. The chain rule is used at the end.
  – herb steinberg
  Jul 26 at 16:19
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The question is phrased in terms of $x$ and $y$, not $x$, $v$, and $y$. Not sure why @Holo edited my original post to appear that way, but I changed it back
  – Atsina
  Jul 25 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I believe you've flipped a negative during the integration
  – Atsina
  Jul 25 at 21:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question can be generalized. Let $f(x)=int_a(x)^b(x)h(c(x)+y)dy$, where $a(x),b(x),c(x),h(x)$ are all known. Then $f(x)=H(c(x)+b(x))-H(c(x)+a(x))$ where $H(x)$ may not be explicitly representable, but $H'(x)=h(x)$. Then $f'(x)=(c'(x)+b'(x))h(c(x)+b(x))-(c'(x)+a'(x))(h(c(x)+a(x))$. The chain rule is used at the end.
  – herb steinberg
  Jul 26 at 16:19
  
  

  
  
  
  

The question is phrased in terms of $x$ and $y$, not $x$, $v$, and $y$. Not sure why @Holo edited my original post to appear that way, but I changed it back
– Atsina
Jul 25 at 21:25

The question is phrased in terms of $x$ and $y$, not $x$, $v$, and $y$. Not sure why @Holo edited my original post to appear that way, but I changed it back
– Atsina
Jul 25 at 21:25

I believe you've flipped a negative during the integration
– Atsina
Jul 25 at 21:35

I believe you've flipped a negative during the integration
– Atsina
Jul 25 at 21:35

This question can be generalized. Let $f(x)=int_a(x)^b(x)h(c(x)+y)dy$, where $a(x),b(x),c(x),h(x)$ are all known. Then $f(x)=H(c(x)+b(x))-H(c(x)+a(x))$ where $H(x)$ may not be explicitly representable, but $H'(x)=h(x)$. Then $f'(x)=(c'(x)+b'(x))h(c(x)+b(x))-(c'(x)+a'(x))(h(c(x)+a(x))$. The chain rule is used at the end.
– herb steinberg
Jul 26 at 16:19

This question can be generalized. Let $f(x)=int_a(x)^b(x)h(c(x)+y)dy$, where $a(x),b(x),c(x),h(x)$ are all known. Then $f(x)=H(c(x)+b(x))-H(c(x)+a(x))$ where $H(x)$ may not be explicitly representable, but $H'(x)=h(x)$. Then $f'(x)=(c'(x)+b'(x))h(c(x)+b(x))-(c'(x)+a'(x))(h(c(x)+a(x))$. The chain rule is used at the end.
– herb steinberg
Jul 26 at 16:19

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up vote
0
down vote

Hint:

Let $$F(x,t)= int_0^x tan(t+y)dy$$

Then we have $$f(x)= F(x,x^2)-F(x,x)$$

NOTE THAT $partial_x F(x,t) = tan(t+x)$ and $partial_t F(x,t) = tan(x+t)- tan(t)$

can you conclude?

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edited Jul 26 at 2:56

answered Jul 26 at 2:46

Guy Fsone

16.8k42671

add a comment | 

up vote
0
down vote

Hint:

Let $$F(x,t)= int_0^x tan(t+y)dy$$

Then we have $$f(x)= F(x,x^2)-F(x,x)$$

NOTE THAT $partial_x F(x,t) = tan(t+x)$ and $partial_t F(x,t) = tan(x+t)- tan(t)$

can you conclude?

share|cite|improve this answer

edited Jul 26 at 2:56

answered Jul 26 at 2:46

Guy Fsone

16.8k42671

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint:

Let $$F(x,t)= int_0^x tan(t+y)dy$$

Then we have $$f(x)= F(x,x^2)-F(x,x)$$

NOTE THAT $partial_x F(x,t) = tan(t+x)$ and $partial_t F(x,t) = tan(x+t)- tan(t)$

can you conclude?

share|cite|improve this answer

edited Jul 26 at 2:56

answered Jul 26 at 2:46

Guy Fsone

16.8k42671

Hint:

Let $$F(x,t)= int_0^x tan(t+y)dy$$

Then we have $$f(x)= F(x,x^2)-F(x,x)$$

NOTE THAT $partial_x F(x,t) = tan(t+x)$ and $partial_t F(x,t) = tan(x+t)- tan(t)$

can you conclude?

share|cite|improve this answer

edited Jul 26 at 2:56

answered Jul 26 at 2:46

Guy Fsone

16.8k42671

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 2:56

edited Jul 26 at 2:56

edited Jul 26 at 2:56

answered Jul 26 at 2:46

Guy Fsone

16.8k42671

answered Jul 26 at 2:46

Guy Fsone

16.8k42671

answered Jul 26 at 2:46

Guy Fsone

16.8k42671

16.8k42671

add a comment | 

add a comment | 

 

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