l'hospital's rule with trig functions


by Shannabel
Tags: functions, lhospital, rule, trig
Shannabel
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#1
Aug12-11, 01:10 PM
P: 74
1. The problem statement, all variables and given/known data
evaluate lim(x->0) (tan^8(t))dt(between 0 and sin^2x)


2. Relevant equations



3. The attempt at a solution
[tan^8(sin^2(x))]/sin^18(x)

my book says to use l'hospital's rule, so i continued with
[8tan^7(sin^2x)*sec^2(sin^2(X))*2sinxcosx]

but my book says i should just have
[tan^8(sin^2x)*2sinxcosx]/18sin^17(x)cos(x)

why did they only partially differentiate tan^8(sin^2(x))?
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Mark44
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#2
Aug12-11, 01:40 PM
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Quote Quote by Shannabel View Post
1. The problem statement, all variables and given/known data
evaluate lim(x->0) (tan^8(t))dt(between 0 and sin^2x)
Is this supposed to be a definite integral? If so, is this the problem?
[tex]\lim_{x \to 0} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]

You can see the LaTeX I used by clicking on the integral.
Quote Quote by Shannabel View Post


2. Relevant equations



3. The attempt at a solution
[tan^8(sin^2(x))]/sin^18(x)
How did you get this?
Quote Quote by Shannabel View Post

my book says to use l'hospital's rule, so i continued with
[8tan^7(sin^2x)*sec^2(sin^2(X))*2sinxcosx]

but my book says i should just have
[tan^8(sin^2x)*2sinxcosx]/18sin^17(x)cos(x)

why did they only partially differentiate tan^8(sin^2(x))?
Shannabel
Shannabel is offline
#3
Aug12-11, 01:46 PM
P: 74
Quote Quote by Mark44 View Post
Is this supposed to be a definite integral? If so, is this the problem?
[tex]\lim_{x \to 0} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]

You can see the LaTeX I used by clicking on the integral.
How did you get this?
no, it should be
[tex]\lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]

Shannabel
Shannabel is offline
#4
Aug12-11, 01:49 PM
P: 74

l'hospital's rule with trig functions


Quote Quote by Shannabel View Post
no, it should be
[tex]\lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]
well i can't make it look right..
but it's 1/[sin^18(x)]
Mark44
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#5
Aug12-11, 01:57 PM
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P: 21,063
Fixed it.
Tip: In LaTeX, if an exponent has more than one character (and that includes pos. or negative sign) put braces {} around it.
Quote Quote by Shannabel View Post
no, it should be
[tex]\lim_{x \to 0} \frac{1}{sin^{18}(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]
Mark44
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#6
Aug12-11, 02:04 PM
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P: 21,063
OK, that makes more sense. L'Hopital's is applicable here.

In answer to your questions, they didn't "partially differentiate" the expression in the integrand - they used the chain rule.

You know (I think) this part of the Fund. Thm. of Calculus involving derivatives and definite integrals.
[tex]\frac{d}{dx} \int_a^x f(t)~dt~= f(x)[/tex]

If the limit of integration contains a function of x, then you need to use the chain rule.
[tex]\frac{d}{dx} \int_a^{g(x)} f(t)~dt~= \frac{d}{du} \int_a^{u} f(t)~dt * \frac{du}{dx}~= f(u) * du/dx[/tex]
SammyS
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#7
Aug12-11, 02:06 PM
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put 18 inside {} → ^{18}
Quote Quote by Shannabel View Post
no, it should be
[tex]\lim_{x \to0} 1/{sin^{18}(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]
Is this the same as: [tex]\lim_{x \to0} \frac{ \int_0^{sin^2(x)} tan^8(t)~dt}{sin^{18}(x)} \ ?[/tex]
Shannabel
Shannabel is offline
#8
Aug12-11, 02:35 PM
P: 74
Quote Quote by Mark44 View Post
OK, that makes more sense. L'Hopital's is applicable here.

In answer to your questions, they didn't "partially differentiate" the expression in the integrand - they used the chain rule.

You know (I think) this part of the Fund. Thm. of Calculus involving derivatives and definite integrals.
[tex]\frac{d}{dx} \int_a^x f(t)~dt~= f(x)[/tex]

If the limit of integration contains a function of x, then you need to use the chain rule.
[tex]\frac{d}{dx} \int_a^{g(x)} f(t)~dt~= \frac{d}{du} \int_a^{u} f(t)~dt * \frac{du}{dx}~= f(u) * du/dx[/tex]
that makes much more sense, thanks!


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