The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.
<h3>What is the derivative of the function g(x) by virtue of the Fundamental theorem of calculus as given in the task content?</h3>
g(x) = Integral; √2 ln(t) dt (with the upper and lower limits e^x and 1 respectively).
Since, it follows from the Fundamental theorem of calculus that given an integral where;
Now, g(x) = Integral f(t) dt with limits a and x, it follows that the differential of g(x);
g'(x) = f(x).
Consequently, the function g'(x) which is to be evaluated in this scenario can be determined as:
g'(x) = = 1
The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.
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Answer:
9 inches
Step-by-step explanation:
4 x 3 = 12
so 3 x 3 = 9
I believe the answer is 21
x^0 + y^0
3^0 + 2^0
1 + 1 =2
your answer : 2
<u>Hint any number with the power of zero is 1</u>
Hope i helped!!! :)