Find the derivatives of the following functions as specified

Question

1. 𝑦 = cos²𝑥 , 𝑦′′′

2. 𝑦 = 𝑥² /2𝑥 + 1 , 𝑦′′′

3. 𝑦 = (𝑥² − 4) ^2/3 , 𝑦′′

4. 𝑦 = 𝑒 ^{𝑥^2} , 𝑦′′′

Answer 1

y= cos^2x = (cosx)^2 (for y=cosx, y'=-sinx)

y' = 2cosx(-sinx) = -2sinxcosx = -sin2x

y" = -2cos2x (y=cosx, y'=-sinx)

y"' = -2(-sin2x)(2) = 4sin2x

or

= 4(2sins2xcos2x) = 8sin2xcos2x

y=x^2/(2x+1)

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

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

y"' = denominator times derivative of numerator minus numerator times derivative of denominator all over denominator squared

= some messy numbers and letters hard to keep straight

y= (x^2 -4)^2/3

y'= (2/3)(x^2-4)^-1/3(2x) = (4x/3)(x^2-4)^-1/3

y' = 1st term times derivative of the 2nd term plus the 2nd term times derivative of the 1st term

=(4x/3)(-1/3)(x^2-4)^-4/3 + (x^2-4)^-1/3(4/3)

y= e^x^2

y' = 2x(e^x^2)

y" = 2x(2x)e^x^2) + e^x^2(2) = 4x^2(e^x^2) + 2e^x^2

= 2e^x^2[2x^2 +1)

with some luck, some of the above don't have mistakes in them.

no guarantees, except double your money back. And double zero = zero.

Plus it's a little hard to read ', ", or "' but that's for 1st 2nd and 3rd derivatives

do most of them, it should answer most of the problems.