4 ms·
No, it is correct. The integral is with respect to x, and the ordinary/partial derivatives are with respect to t. Written out fully, the derivative computation
by owalt 10mo ago
No, it is correct. The integral is with respect to x, and the ordinary/partial derivatives are with respect to t. Written out fully, the derivative computation is
d/dt (x^t - 1)/ln(x) = d/dt [exp(ln(x)t) - 1]/ln(x)
= ln(x)exp(ln(x)t)/ln(x)
= exp(ln(x)t)
= x^t.
Edit: d/dt exp(ln(x)t) = ln(x)exp(ln(x)t) by the chain rule, while d/dt (1/ln(x)) = 0 since the expression is constant with respect to t.
There are convergence considerations that were not discussed in the blog post, but the computations seem to be correct.
- impossiblefork 10mo agoAh, yes. I don't understand how I differentiated with respect to x instead of t, but...