In [15]: bcdata.with_column("daily change",h-1) Out [15]: Date Price Open High Low Vol. daily change 659.6 37.12K 724.4 38.15K 525.9 98.85K 1572.8 Jun 26, 2023 30,248.1 30,466.3 30,645.9 29,986.3 58.32K Jun 25, 2023 30,465.3 30,533.6 31,040.2 30,315.8 Jun 24, 2023 30,533.6 30,680.7 30,795.3 30,269.4 Jun 23, 2023 30,679.4 29,890.2 31,395.4 29,822.6 Jun 22, 2023 29,890.5 29,992.8 30,497.8 29,590.4 79.22K Jun 21, 2023 29,996.9 28,307.7 30,769.5 28,270.5 143.49K Jun 20, 2023 28,307.7 26,845.9 28,393.0 26,665.5 100.55K Jun 19, 2023 26,845.9 26,339.7 27,029.7 26,295.1 46.45K Jun 18, 2023 26,339.7 26,515.0 26,679.3 26,290.6 27.31K Jun 17, 2023 26,515.0 26,341.3 26,767.3 26,183.5 35.86K 907.4 2499 1727.5 734.6 388.7 583.8 .. (22 rows omitted) Question 1 Write a function that accepts a date as input and returns the corresponding day of the week for that specific date. You may need to import the date or time modules. For instance, if the input is June 29, 2023, the output should be Thursday. In [5] #your code here def