$F(x)=int_{pi}^{x} frac{e^t}{3+sin t} dt$ Find $(F^{-1})'(0)$
Let $F$ be a function: $F(x)=int_{pi}^{x} frac{e^t}{3+sin t} dt$
Find $(F^{-1})'(0)$.
What I've been doing:
What I tried to do was to find $F'(x)$ and $F^{-1}(0)$ such that I could use $frac{1}{F'(F^{-1}(x))}$
I found that $F'(x)=frac{e^x}{3+sin x}$, however when I try to integrate the original function I can't. I tried doing it by substitution, with $u=e^x$ or $u=3+sin x$.
Is there something I'm missing? Do I need to integrate when trying to find $(F^{-1})'(0)$?
calculus real-analysis integration derivatives definite-integrals
asked Nov 24 at 22:16
parishilton
19310
add a comment |
Let $F$ be a function: $F(x)=int_{pi}^{x} frac{e^t}{3+sin t} dt$
Find $(F^{-1})'(0)$.
What I've been doing:
What I tried to do was to find $F'(x)$ and $F^{-1}(0)$ such that I could use $frac{1}{F'(F^{-1}(x))}$
I found that $F'(x)=frac{e^x}{3+sin x}$, however when I try to integrate the original function I can't. I tried doing it by substitution, with $u=e^x$ or $u=3+sin x$.
Is there something I'm missing? Do I need to integrate when trying to find $(F^{-1})'(0)$?
calculus real-analysis integration derivatives definite-integrals
asked Nov 24 at 22:16
parishilton
19310
add a comment |
Let $F$ be a function: $F(x)=int_{pi}^{x} frac{e^t}{3+sin t} dt$
Find $(F^{-1})'(0)$.
What I've been doing:
What I tried to do was to find $F'(x)$ and $F^{-1}(0)$ such that I could use $frac{1}{F'(F^{-1}(x))}$
I found that $F'(x)=frac{e^x}{3+sin x}$, however when I try to integrate the original function I can't. I tried doing it by substitution, with $u=e^x$ or $u=3+sin x$.
Is there something I'm missing? Do I need to integrate when trying to find $(F^{-1})'(0)$?
calculus real-analysis integration derivatives definite-integrals
asked Nov 24 at 22:16
parishilton
19310
Let $F$ be a function: $F(x)=int_{pi}^{x} frac{e^t}{3+sin t} dt$
Find $(F^{-1})'(0)$.
What I've been doing:
What I tried to do was to find $F'(x)$ and $F^{-1}(0)$ such that I could use $frac{1}{F'(F^{-1}(x))}$
I found that $F'(x)=frac{e^x}{3+sin x}$, however when I try to integrate the original function I can't. I tried doing it by substitution, with $u=e^x$ or $u=3+sin x$.
Is there something I'm missing? Do I need to integrate when trying to find $(F^{-1})'(0)$?
calculus real-analysis integration derivatives definite-integrals
calculus real-analysis integration derivatives definite-integrals
asked Nov 24 at 22:16
parishilton
19310
asked Nov 24 at 22:16
parishilton
19310
asked Nov 24 at 22:16
parishilton
19310
asked Nov 24 at 22:16
parishilton
19310
asked Nov 24 at 22:16
parishilton
19310
19310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
All you need to do isbegin{align}(F^{-1})'(0)&=frac1{F'bigl(F^{-1}(0)bigr)}\&=frac1{F'(pi)}\&=frac{3^pi}{1+sinpi}\&=3^pi.end{align}
answered Nov 24 at 22:19
José Carlos Santos
150k22120221
Sorry for maybe silly question, but how did you get the $F^{-1}(0) = pi$?.
– Makina
Nov 24 at 22:37
2
That's because $F(pi)=0$.
– José Carlos Santos
Nov 24 at 22:38
1
Oh ye, damn it, so simple ... I thought there was some theory behind it. Thank you for clarification.
– Makina
Nov 24 at 22:39
Didn't know it was that easy! Thank you!!
– parishilton
Nov 24 at 22:52
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
All you need to do isbegin{align}(F^{-1})'(0)&=frac1{F'bigl(F^{-1}(0)bigr)}\&=frac1{F'(pi)}\&=frac{3^pi}{1+sinpi}\&=3^pi.end{align}
answered Nov 24 at 22:19
José Carlos Santos
150k22120221
Sorry for maybe silly question, but how did you get the $F^{-1}(0) = pi$?.
– Makina
Nov 24 at 22:37
2
That's because $F(pi)=0$.
– José Carlos Santos
Nov 24 at 22:38
1
Oh ye, damn it, so simple ... I thought there was some theory behind it. Thank you for clarification.
– Makina
Nov 24 at 22:39
Didn't know it was that easy! Thank you!!
– parishilton
Nov 24 at 22:52
add a comment |
All you need to do isbegin{align}(F^{-1})'(0)&=frac1{F'bigl(F^{-1}(0)bigr)}\&=frac1{F'(pi)}\&=frac{3^pi}{1+sinpi}\&=3^pi.end{align}
answered Nov 24 at 22:19
José Carlos Santos
150k22120221
Sorry for maybe silly question, but how did you get the $F^{-1}(0) = pi$?.
– Makina
Nov 24 at 22:37
2
That's because $F(pi)=0$.
– José Carlos Santos
Nov 24 at 22:38
1
Oh ye, damn it, so simple ... I thought there was some theory behind it. Thank you for clarification.
– Makina
Nov 24 at 22:39
Didn't know it was that easy! Thank you!!
– parishilton
Nov 24 at 22:52
add a comment |
All you need to do isbegin{align}(F^{-1})'(0)&=frac1{F'bigl(F^{-1}(0)bigr)}\&=frac1{F'(pi)}\&=frac{3^pi}{1+sinpi}\&=3^pi.end{align}
answered Nov 24 at 22:19
José Carlos Santos
150k22120221
All you need to do isbegin{align}(F^{-1})'(0)&=frac1{F'bigl(F^{-1}(0)bigr)}\&=frac1{F'(pi)}\&=frac{3^pi}{1+sinpi}\&=3^pi.end{align}
answered Nov 24 at 22:19
José Carlos Santos
150k22120221
answered Nov 24 at 22:19
José Carlos Santos
150k22120221
answered Nov 24 at 22:19
José Carlos Santos
150k22120221
answered Nov 24 at 22:19
José Carlos Santos
150k22120221
150k22120221
Sorry for maybe silly question, but how did you get the $F^{-1}(0) = pi$?.
– Makina
Nov 24 at 22:37
2
That's because $F(pi)=0$.
– José Carlos Santos
Nov 24 at 22:38
1
Oh ye, damn it, so simple ... I thought there was some theory behind it. Thank you for clarification.
– Makina
Nov 24 at 22:39
Didn't know it was that easy! Thank you!!
– parishilton
Nov 24 at 22:52
add a comment |
Sorry for maybe silly question, but how did you get the $F^{-1}(0) = pi$?.
– Makina
Nov 24 at 22:37
2
That's because $F(pi)=0$.
– José Carlos Santos
Nov 24 at 22:38
1
Oh ye, damn it, so simple ... I thought there was some theory behind it. Thank you for clarification.
– Makina
Nov 24 at 22:39
Didn't know it was that easy! Thank you!!
– parishilton
Nov 24 at 22:52
Sorry for maybe silly question, but how did you get the $F^{-1}(0) = pi$?.
– Makina
Nov 24 at 22:37
Sorry for maybe silly question, but how did you get the $F^{-1}(0) = pi$?.
– Makina
Nov 24 at 22:37
2
2
That's because $F(pi)=0$.
– José Carlos Santos
Nov 24 at 22:38
That's because $F(pi)=0$.
– José Carlos Santos
Nov 24 at 22:38
1
1
Oh ye, damn it, so simple ... I thought there was some theory behind it. Thank you for clarification.
– Makina
Nov 24 at 22:39
Oh ye, damn it, so simple ... I thought there was some theory behind it. Thank you for clarification.
– Makina
Nov 24 at 22:39
Didn't know it was that easy! Thank you!!
– parishilton
Nov 24 at 22:52
Didn't know it was that easy! Thank you!!
– parishilton
Nov 24 at 22:52
add a comment |
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