Question

Just got through with derivatives, power rule, and chain rule.

I am supposed to find the first and second derivatives of the function.

y = 4x/√x+1 and y = cos^2(x)

y' = [(x+1)^(1/2)(4)]-[(4x)(1/2(x+1)^(-1/2))(1)]
y' = [4(x+1)^(1/2)-4x]/[2(x+1)^(1/2)(x+1)]
y' = (2-4x)/(x+1)
y" = [(x+1)(-4)]-[(2-4x)(1)]/(x+1)^2
y" = -6/(x+1)^2

y' = (cosx)^2
y' = 2(cosx)(-sinx)
y" = (-2cosx)(cosx)+(-sinx)(0)(-sinx)
y" = -2cos^2(x)

I feel like I am struggling with the chain rule and am just practicing

Answer 1

y = 4 x (x+1)^-.5
y' = 4x[-.5(x+1)^-1.5] +(x+1)^-.5[4}
= -2x /[(x+1)(x+1)^.5] + 4/(x+1)^.5
= -2x /[(x+1)(x+1)^.5] + 4(x+1)/[(x+1)](x+1)^.5
= (2x+4)(x+1)^-1.5
now do second

y = cos^2 x
y' = 2 cos x (-sin x) agree
but then
y" = 2 [ cos x (-cos x) -sin x(-sin x) ]
=2 [ - cos^2 x + sin^2 x]

Answer 2

y" = (2x+4)(-1.5(x+1)^-2.5) + (x+1)^-1.5(2)
y" = (x+1)^-1.5[2+(2x+4)(-3/2(x+1))]
y" = (x+1)^-1.5[2+(-6(x+1)/2(x+1))]
y" = (x+1)^-1.5[2+(-3)]
y" = -(x+1)^-1.5

Also, thank you for your time!!