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Derivatives and Functions
Derivatives and Functions
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Explore questions about finding derivatives and evaluating functions in calculus
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1
Find ๐โฒ(๐ฅ) for ๐(๐ฅ) = (3๐ฅยฒ + 5๐ฅ)(4๐ฅยฒ + 3๐ฅ)
24๐ฅยณ + 39๐ฅยฒ + 15๐ฅ
12๐ฅยณ + 39๐ฅยฒ + 15๐ฅ
12๐ฅยณ + 39๐ฅยฒ + 30๐ฅ
24๐ฅยณ + 39๐ฅยฒ + 30๐ฅ
2
Find ๐ฆโฒ for ๐ฆ = tan(5๐ฅยณ)
15๐ฅsecยฒ(5๐ฅยณ)
5๐ฅsecยฒ(5๐ฅยณ)
5๐ฅยฒsecยฒ(5๐ฅยณ)
15๐ฅยฒsecยฒ(5๐ฅยณ)
3
Find ๐ฆโฒ for ๐ฆ = (5๐ฅยฒ + 3๐ฅ)โด / (๐ฅยฒ + 4)
10๐ฅ(5๐ฅยฒ + 3๐ฅ)ยฒ / (๐ฅยฒ + 4)
20๐ฅ(5๐ฅยฒ + 3๐ฅ)ยฒ / (๐ฅยฒ + 4)
20๐ฅ(5๐ฅยฒ + 3๐ฅ)ยณ / (๐ฅยฒ + 4)2
2x3(5x+3)3(15x3+3x2+80x+24)/(x2+4)2
4
Find ๐โฒ(๐ฅ) for ๐(๐ฅ) = sin(2๐ฅยณ)
6๐ฅยฒcos(2๐ฅยณ)
6๐ฅcos(2๐ฅยณ)
2๐ฅยฒcos(2๐ฅยณ)
2๐ฅcos(2๐ฅยณ)
5
Find [๐(๐ฅ) โ ๐(๐ฅ)]โฒ for ๐(๐ฅ) = 7๐ฅยณ โ 6๐ฅยฒ and ๐(๐ฅ) = 6๐ฅยฒ + 8๐ฅ
21๐ฅยฒ โ 20๐ฅ
7๐ฅยฒ + 20๐ฅ
21๐ฅยฒ + 20๐ฅ
7๐ฅยฒ โ 20๐ฅ
6
Find ๐โฒ(๐ฅ) for ๐(๐ฅ) = 2๐ฅโปยณ / (5๐ฅยฒ โ 3๐ฅ + 2)
25x2-24x+12/ (5๐ฅยฒ โ 3๐ฅ + 2)2
6๐ฅโปยณ / (5๐ฅยฒ โ 3๐ฅ + 2)
50x2-24x+12/x4(5x2-3x+2)2
-6๐ฅโปยณ / (5๐ฅยฒ โ 3๐ฅ + 2)
7
Find ๐โฒ(๐ฅ) for ๐(๐ฅ) = 6๐ฅยฒ โ 5๐ฅ + 7
6๐ฅ โ 5
12๐ฅ + 7
6๐ฅยฒ โ 5
12๐ฅ โ 5
8
What is 4๐โฒ(๐ฅ) for ๐(๐ฅ) = 2๐ฅโด โ 3๐ฅยณ + 5๐ฅยฒ + 4?
32๐ฅยณ โ 36๐ฅยฒ + 40๐ฅ
8๐ฅยณ โ 9๐ฅยฒ + 10๐ฅ
8๐ฅยณ โ 9๐ฅยฒ + 20๐ฅ
32๐ฅยณ โ 36๐ฅยฒ + 20๐ฅ
9
Find ๐ฆโฒ for ๐ฆ = (๐ฅยณ + 3)โต
15๐ฅ(๐ฅยณ + 3)โด
5๐ฅยฒ(๐ฅยณ + 3)โด
5๐ฅ(๐ฅยณ + 3)โด
15๐ฅยฒ(๐ฅยณ + 3)โด
10
Find (๐๐)โฒ for ๐(๐ฅ) = ๐ฅ + 3 and ๐(๐ฅ) = 2๐ฅ โ 5
2๐ฅ โ 3
2๐ฅ โ 1
2๐ฅ + 3
2๐ฅ + 1