- #1
scorpa
- 367
- 1
Hello Everyone,
I'm doing some questions on higher derivatives, and they should be easy but I am worried that my answers are not quite right. There just seems to be something 'off' about them. Anyway here are the questions and my answers.
1) Find the first and second derivatives of y= (4x)/squareroot(x+1)
y' =(squareroot(x+1))(4)-(4x)(1/2)((x+1)^-1/2))/(x+1)
y'=(4squareroot(x+1)-(2x(x+1)^-1/2)/(x+1)
Ok, so that was my first derivative and I think it is ok, now for the second:
y''=((x+1)[(2(x+1)^-1/2)-(2(x+1)^-1/2)-(x(x+1)^-3/2)] -4squareroot(x+1)+2x(x+1)^-1/2)/(x+1)^2
I hope that doesn't look to confusing but I have a bad feeling it is.
2) Find the first and second derivatives of x^4+y^4=a^4
My biggest issue with this one is that I'm unsure of how to deal with the a, I just treated it like the x, and used implicit differentiation:
x^4 + y^4 = a^4
4x^3 + 4y^3 y' = 4a^3
y' =(4a^3 - 4x^3)/4y^3
y' = (a^3 -x^2)/y^3
y'' = y^3(3a^2-3x^2)-3y^2(a^3-x^3)/(y^6)
y'' = 3y^2 (ya^2-yx^2-a^3+x^3)/y^6
y'' = (3(ya^2-yx^2-a^3+x^3))/y^4
Well there it is, hope what I've shown makes sense, I know it is hard to read on a computer. Thanks for any help guys, I really appreciate your time and effort.
I'm doing some questions on higher derivatives, and they should be easy but I am worried that my answers are not quite right. There just seems to be something 'off' about them. Anyway here are the questions and my answers.
1) Find the first and second derivatives of y= (4x)/squareroot(x+1)
y' =(squareroot(x+1))(4)-(4x)(1/2)((x+1)^-1/2))/(x+1)
y'=(4squareroot(x+1)-(2x(x+1)^-1/2)/(x+1)
Ok, so that was my first derivative and I think it is ok, now for the second:
y''=((x+1)[(2(x+1)^-1/2)-(2(x+1)^-1/2)-(x(x+1)^-3/2)] -4squareroot(x+1)+2x(x+1)^-1/2)/(x+1)^2
I hope that doesn't look to confusing but I have a bad feeling it is.
2) Find the first and second derivatives of x^4+y^4=a^4
My biggest issue with this one is that I'm unsure of how to deal with the a, I just treated it like the x, and used implicit differentiation:
x^4 + y^4 = a^4
4x^3 + 4y^3 y' = 4a^3
y' =(4a^3 - 4x^3)/4y^3
y' = (a^3 -x^2)/y^3
y'' = y^3(3a^2-3x^2)-3y^2(a^3-x^3)/(y^6)
y'' = 3y^2 (ya^2-yx^2-a^3+x^3)/y^6
y'' = (3(ya^2-yx^2-a^3+x^3))/y^4
Well there it is, hope what I've shown makes sense, I know it is hard to read on a computer. Thanks for any help guys, I really appreciate your time and effort.