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I have tha answer can someone please just check me... Find the derivative of p(x)=cos(ln(3x-1))
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you got what?
I got \[\frac{ 3\sin(\ln(3x-1)) }{ 3x-1 }\]
again you lost a negative sign somewhere, the derivative of cosine is negative sine
p(x) = cos(ln(3x-1)) p ' (x) = d/dx[ cos(ln(3x-1)) ] p ' (x) = -sin(ln(3x-1))*d/dx[ln(3x-1)] p ' (x) = -sin(ln(3x-1))/(3x-1)*d/dx[3x-1] p ' (x) = -sin(ln(3x-1))/(3x-1)*3 p ' (x) = -3sin(ln(3x-1))/(3x-1)
other than that, everything else is correct
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ok so its just a negative on the top then
yep
thanks
np
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