How do I find the first and second derivative of this function?

Question

y = xsqrt(2-x^2)

Answer 1

first derivative.

use product rule:
d/dx (uv) = u'v + v'u

y' = d/dx (x) * √(2 - x^2) + d/dx (√(2 - x^2)) (x)
y' = √(2 - x^2) + d/dx (√(2 - x^2)) x

chain rule:
let u = 2 - x^2
then d/du (√u) = 1/(2√u)

d/dx (2 - x^2) = -2x

y' = √(2 - x^2) + (-2x) 1/[2√(2-x^2)] * x
y' = √(2 - x^2) - x^2/√(2-x^2)

multiply both top and bottom to the first term by √(2-x^2)
y' = (2-x^2) / √(2-x^2) - x^2/√(2-x^2)
y' = (2 - x^2 - x^2) / √(2 - x^2)
y' = (2 - 2x^2) / √(2 - x^2)
y' = 2(1-x^2) / √(2-x^2) <== answer
OR y' = -2 (x^2 - 1) / √(2-x^2) <== answer
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second derivative is the deivative ofthe first derivative
use quotien rule:
d/dx (u/v) = (u'v - v'u)/v^2

y" = 2 [d/dx (1-x^2) √(2-x^2) - d/dx (√(2-x^2) (1-x^2)] / (2-x^2)

in first derivative, we already found the the derivate of
√(2-x^2) is - x/√(2-x^2)

y" = 2[ (-2x) √(2-x^2) - (-x)/√(2-x^2) (1 - x^2)] / (2-x^2)

y" = 2[ (-2x)√(2-x^2) + (x - x^3)/√(2-x^2)] / (2-x^2)

multiply √(2-x^2) for both top and bottom on the numerator

y" = 2 [ (-2x) (2-x^2)/(2-x^2) + (x - x^3)/√(2-x^2)] / (2-x^2)

y" = 2 [ (-4x + 2x^3 + x - x^3) / √(2-x^2)] / (2 - x^2)

y" = 2 (x^3 - 3x) / [√(2-x^2)] * (2 - x^2)]

y' = 2x (x^2 - 3) / (2 - x^2)^(3/2) <== answer


Rec

Answer 2

use the chain rule to find the differential of sqrt(2-x^2)
then use the product rule for xsqrt(2-x^2)

Answer 3

i do it for sqrt as power function
y=x*(2-x^2)^(1/2)
for the multiply function derivative is d of 1st * 2nd part and d of 2nd to 1st
so:
y'=1*(2-x^2)(1/2)+(1/2)*(2-x^2)^(-1/2)*(-2*x)*x
and do it like this for second part

Answer 4

y = x (2 - x²) ^(1/2)
dy/dx is given by:-
(2- x²)^(1/2) + (1/2)(2 - x²)^(-1/2)(- 2x)
(2 - x²)^(1/2) - (2 - x²)^(-1/2) (x)
(2 - x²)^(-1/2) [ (2 - x²) - x ]

d²y/dx² is given by:-
(-1/2)(2 - x²)^(-3/2)(- 2x) [ 2 - x² - x ] +
(- 2x - 1)(2 - x²)^(-1/2)

(2 - x²)^(-3/2) [ x (2 - x² - x) + (-2x - 1)(2 - x²) ]
(2 - x²)^(-3/2) [2x - x³ - x² + (- 4x + 2x³ - 2 + x²)]
(2 - x²)^(-3/2) [ x³ - 2x - 2 ]

Difficult to type---hope I have not made any errors.