Fundamental Theorem of Calculus to find Derivative

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SUMMARY

The discussion centers on the application of the Fundamental Theorem of Calculus (FTC) to find the derivative of the function g(x) = √x ∫ sin(x) ln(t) (cos(t)/t) dt. The correct derivative is established as g'(x) = ln(x) cos(x) / x. Participants clarify that g(x) should be defined as g(x) = √x sin(x) ∫_{a}^x (ln(t) cos(t)/t) dt, necessitating the use of the product rule for differentiation.

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Use the Fundamental Theorem of Calculus to find the derivative of the function



g(x) = [tex]\sqrt{x}\int sinx[/tex] Ln(t) [tex]\frac{cos(t)}{t}[/tex] dt



g'(x) = lnx cosx / x. By integrating this function, you receive the function g(x). Then by differentiating g(x) you receive g'(x) which is what is given, according to FTCI.


I was told I got this completely wrong. (out of 5) So looking for some help. Thanks in advance.



 
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I do not understand something. Could you possibly rewrite the original problem?
Is it:

[tex]g(x)=\sqrt{x}\int{sin(x)dx} *\int{\frac{ln(t)*cos(t)}{t}dt}[/tex]


Regards.
 
As stated, the function [tex]g(x)[/tex] doesn't look quite right. Is it supposed to be defined such that,

[tex]g(x) = \sqrt x sin(x) \int_{a}^x \frac{ln(t)cos(t)}{t} \, dt[/tex]

If so, then use the product rule of differentiation because [tex]g(x)[/tex] can be defined in terms of the product of two functions [tex]u[/tex] and [tex]v[/tex] where,

[tex]u = \sqrt x sin(x)[/tex]

[tex]v = \int_{a}^x \frac{ln(t)cos(t)}{t} \, dt[/tex]
 

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