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I have the following problem:

Let $f(x) = \int_{e^{3x}}^{2x^2}\ln(t)dt$. Evaluate $f'(x)$ at $x = 1$

My approach is to directly calculate the integral using integration by parts, and then just take the derivative of the result and compute it at $x=1$.

I feel like there may be a quicker way using the Fundamental theorem of Calculus but I am not sure. Is my approach correct and is there another approach to this question? Thank you!

Gary
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sucksatmath
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    Use Leibnitz's formula. – Nimu Basak Aug 24 '23 at 02:13
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    Write it as $F(2x^2 ) - F({\rm e}^{3x} )$ with $F(x) = \int_{\rm 1}^x {\ln (t){\rm d}t}$, then use the chain rule and the FTC. – Gary Aug 24 '23 at 02:14
  • https://math.stackexchange.com/questions/781398/derivative-of-integral-with-variable-bounds See formula in general for this in first answer. – coffeemath Aug 24 '23 at 02:15
  • Based on calculus on many variables and linearity of derivations, consider each occurrence of x as a different variable $x_i$, do the differntial and add all coefficients $$d/dx f(x,x) = ( df(x,y)/dx + df(x,y)/dy ) /. y\to x $$ – Roland F Aug 24 '23 at 05:32

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