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find the derivative of F(x)=integral from x to 4 of (sin(t^(4)))dt
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\[\int_x^4sin(t^4)dt?\]
well what is \[\frac{d}{dx}\int\]
yeah that^
use the fundamental theorem of calculus
\[f(x)=\frac{d}{dx}F(x)\]
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\[F(x)=\int_x^4sin(t^4)dt\]
Okay, gimme a sec.
I find F(x) very vague for this i think it should be \[F(x)=\int_x^4 sin(t^4)=F(4)-F(x) \] \[\frac{d}{dx}(F(4)-F(x))=0-F'(x)(1)=-f(x)\]
the one inside is F(t) for x whereas the outside is just F(x)
so it ends up being \[f(t)=sin(t^4)\] \[f(x)=f(t)|_0^x\] you can say i guess
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\[-sin(x^4)\]
make sense?
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