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Homework Statement
evaluate lim(x->0) (tan^8(t))dt(between 0 and sin^2x)
Homework Equations
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))?