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How would I go about taking the double derivative of the function F(x) = integral[1, 3x] ln(t^2)dt? The only problem I have is with the initial derivative.
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\(\Large \dfrac{d}{dx}\int \limits_{a}^{g(x)}h(x)dx = h(g(x))g'(x) \)
So F'(x) would be 3ln((3x)^2)?
yes!
you can simplify that further
3ln(9(x^2))?
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\(\ln A^2 = 2 \ln A\)
6ln(3x)? or would it be 6ln(9x)?
Oh, 6ln(3x).
yes
Thank you so much for the help.
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need help in finding its derivative? whats the final answer you got? just for verification
6/x
\(\huge \checkmark \)
Thanks.
welcome ^_^
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