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Calculus Derivatives help?
I'm having trouble finding the first derivative of these two problems, any help would be much appreciated.
a. 5x^3-3x^5-3tan^6(pi/12)^2
b. (x^2*sqrt(x^3+1))/(x-2)
3 Answers
- 2 years agoFavorite Answer
What in the world is 3 * tan^6(pi/12)^2? Is it 3 * tan((pi/12)^2)^6? Either way, without a variable present, you could just call it a constant C
5x^3 - 3x^5 - C
That's easy enough to derive
15x^2 - 15x^4
x^2 * sqrt(x^3 + 1) / (x - 2)
u = x^2 * sqrt(x^3 + 1)
u^2 = x^4 * (x^3 + 1)
u^2 = x^7 + x^4
2u * du = 7x^6 * dx + 4x^3 * dx
du/dx = (7x^6 + 4x^3) / (2u)
u' = x^3 * (7x^3 + 4) / (2 * x^2 * sqrt(x^3 + 1))
u' = x * (7x^3 + 4) / (2 * sqrt(x^3 + 1))
v = x - 2
v' = 1
d(u/v) = (v * u' - u * v') / v^2
((x - 2) * x * (7x^3 + 4) / (2 * sqrt(x^3 + 1)) - x^2 * sqrt(x^3 + 1) * 1) / (x - 2)^2 =>
(1/sqrt(x^3 + 1)) * ((x - 2) * x * (7x^3 + 4) * sqrt(x^3 + 1) - x^2 * (x^3 + 1)) / (x - 2)^2 =>
(x * (x - 2) * (7x^3 + 4) * sqrt(x^3 + 1) / 2 - x^2 * (x^3 + 1)) / ((x - 2)^2 * sqrt(x^3 + 1)) =>
(x * (x - 2) * (7x^3 + 4) * sqrt(x^3 + 1) - 2 * x^2 * (x^3 + 1)) / (2 * (x - 2)^2 * sqrt(x^3 + 1))
- Anonymous2 years ago
That question is too hard for people on here you need to go ask a university professor


