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OpenStudy (anonymous):
f(x)=\int_1^{3x^5} \ln(t^2+1) \, dt
OpenStudy (anonymous):
\[f(x)=\int_1^{3x^5} \ln(t^2+1) \, dt\]like that?
OpenStudy (anonymous):
is the question find the derivative?
OpenStudy (anonymous):
yes it is,
OpenStudy (anonymous):
how'd i guess?
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OpenStudy (anonymous):
too hard to integrate
OpenStudy (anonymous):
use the chain rule. the derivative of the integral is the integrand, so replace t by
\[3x^5\] and then multiply the answer by
\[15x^4\]
OpenStudy (anonymous):
yeah im having trouble with these. Ill try it sat!
OpenStudy (anonymous):
\[f'(x)=\ln((3x^5)^2+1)\times 15x^4\]
OpenStudy (anonymous):
i should have let you do it on your own, but once you see it you should realize how easy it is. it is clear what i did yes?
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OpenStudy (anonymous):
if you need a word of explanation let me know
OpenStudy (anonymous):
i understand chain rule for ln (function within function). where did the 15x4 come from?
OpenStudy (anonymous):
nvm 3x^5 derivative
OpenStudy (anonymous):
15x^4 \ln(9x^{10}+1) ?
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
it is the chain rule and you have a composite function. don't forget that
\[F(x)=\int_a^x f(t)dt\] is a function of "x" so if you do not have an x in the upper limit, but rather something else, you have a composition like
\[F(g(x))=\int_a^{g(x)}f(t)dt\] and so you need the chain rule