D e^(b*t*ln(t)) + ln(x) respect to t
t^(b*t)*(ln(t)*b)+b + 1/x
No. For the exponential function, the chain rule looks like this: d/dt(eu) = eu*du/dt.
Also, did you mean the ln(x) term to be ln(t)?
nope. It is ln(x)
And are you supposed to find the derivative with respect to t or is it the partial derivative with respect to t?
Your D' notation doesn't mean anything, so I can't tell what you are trying to do.
derivative respect to t <:
= e^(b*t*ln(t) * (b*ln(t)+b)+1/x
1) d/dt(ln(x)) = 0. As far as t is concerned, x is a constant, so ln(x) is a constant, so d/dt(ln(x)) is 0.
2) b*t*1/t can be simplified.
Also, it would be good for you to be more explanatory in what you are doing. For this problem, that would mean indicating that you are taking the derivative of something, with respect to some variable, and then showing what you arrived at.
For this problem, what I'm talking about is this:
d/dt[e^(b*t*ln(t)) + ln(x)] = exp(b*t*ln(t)*(b*ln(t)+(b*t*1/t)+1/x
Note that I copied your result verbatim, and the two points above still apply.
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