Find the second derivative of f(x)=(3-x^2)^5?

Answer 1

f(x) = (3-x^2)^5

f'(x) = 5(3 - x^2)^4 * d/dx(3 - x^2) = -10x (3- x^2)^4

Using the chain rule

f"(x) = -10x d/dx((3- x^2)^4)) + (3- x^2)^4 * d/dx(-10x)
f"(x) = -10x *4[(3-x^2)^3] * (-2x) -10(3- x^2)^4

f"(x) = 80x^2 * (3 - x^2)^3 -10 (3 -x^2)^4
f"(x) = (3 - x^2)^3 [ 80x^2 - 10(3 - x^2)] =
= (90x^2 - 30) (3 - x^2)^3

and finally

f"(x) = 30 (3x^2 - 1) (3 - x^2)^3

Answer 2

f(x) = (3- x^2)^5
f(x) is also = y
taking log on bothsides
logy= log(3- x^2)^5

logy =5 . log(3- x^2)
differenciating
y'/y =5./(3 -x^2 ) -2x
y' = -10x/(3 -x^2) (3- x^2)^5
y' = -10x .(3- x^2)^4 ...... (1)
diff again
using chain & product rule
y''= -10x .d/dx(3- x^2)^4 + (3- x^2)^4 .-10 .d/dx(-x)
y''= -10x .4(3- x^2)^3.d/dx(3 - x^2) + (3- x^2)^4.10
y'' = - 40x.(3- x^2)^3.-2x + (3- x^2)^4.10
= +80x (3- x^2)^3 + (3- x^2)^4.10
y'' = +80x (3- x^2)^3 - y'/x (from (1) )

Answer 3

Chain, power, and product rules here:

f'(x)=5((3-x^2)^4)(-2x)=-10x(3-x^2)^4

f''(x)=-10(4x(3-x^2)^3 (-2x)+(3-x^2)^4)

=-10(3-x^2)^3 (3-x^2-8x^2)= -30(3-x^2)^3(1-3x^2)=30(3-x^2)^3(3x^2-1)

Answer 4

f `(x) = 5.(3 - x²)^4.(-2x)
f `(x) = (-10 x).(3 - x²)^4
f " (x) = (-10).(3 - x²)^4 + 4.(3 - x²)³.(-2x).(-10x)
f "(x) = (-10).(3 - x²)^4 + 80x².(3 - x²)³
f " (x) = (-10).(3 - x²)³ [ (3 - x²) - 8x² ]
f "(x) = (-10).(3 - x²)³ [ 3 - 9x² ]
f "(x) = (-30).(3 - x²).(1 - 3x²)

Answer 5

f''(x) = -10*(3-x^2)^4 + 80*(x^2)*(3-x^2)^3
= 30(3-x^2)^3*(3x^2-1)

Answer 6

df(x)/dx=5*((3-x^2)^4)(-2x)
=-10x(3-x^2)^4

d(df(x)/dx)/dx=-10*([(3-x^2)^4]+x*4([(3-x^2)^3][-2x]))
=-10*([(3-x^2)^4]-(8x^2)[(3-x^2)^3])
=-10[(3-x^2)^4]+(80x^2)[(3-x^2)^3]