Different result by hand-differentiaion vs MATLAB diff() function.

I differentiated the equation, ( K=(y^2 + 2*y + 2)/(y + 1)^2 ), by hand-differentiation and got result, ( -2/(y+1) ) while the result with MATLAB function, ( diff(K, y) ) was, (2*y + 2)/(y + 1)^2 - (2*(y^2 + 2*y + 2))/(y + 1)^3

I don't know why was it and what was the wrong I did.

Please help. Thank you.

Answer 1

[Roger Stafford]

Your "hand-differentiation" is not correct! The 'diff' answer is correct but can be simplified to an equivalent -2/(y+1)^3. (Perhaps the latter is what you meant to write?)

[yf]

Hi Mr. Roger Stafford, Thank you for your answer. But I followed the procedure described in the

 'Solutions Manual - Control Systems Engineering by Norman S. Nise ed 6.pdf'

and got the same result ( please refer to the attached file, image1.jpg, for reference.

That is why I cannot understand. Please help. Thanks again.

[Roger Stafford]

Here is the differentiation computation:

K = (y^2 + 2*y + 2)/(y + 1)^2
dK/dy = (d(y^2 + 2*y + 2)/dy)/(y+1)^2 + (d((y+1)^-2)/dy)*(y^2 + 2*y + 2)
  = (2*y+2)/(y+1)^2 - 2*((y+1)^-3)*(y^2 + 2*y + 2)
  = ((2*y+2)*(y+1)-2*(y^2 + 2*y + 2))/(y+1)^3
  = ((2*y^2+4*y+2)-(2*y^2+4*y+4)) / (y+1)^3
  = -2/(y+1)^3

Somehow you lost the cube power in that y+1 denominator.

[Stephen23]

yf's "Answer" moved here:

Hi Mr. Roger Stafford,

Thank you for your answer. I will think about it why is the different result in between the book and from the MATLAB. Thanks once again.

[Stephen23]

@yf: considering that Roger Stafford helped you to resolve this issue then you should accept this answer: this is the easiest way for you to show your appreciation to the volunteers who help you on this forum.

[Stephen23]

yf's "Answer" moved here:

Hi Mr. Roger Stafford,

I found the problem now. the denominator in the solution manual is actually,

(y^2 + 2*y + 1)^2 = ( (y + 1)^2 )^2

hence is,

(2*y + 2)/(y + 1)^2 - (2*(y^2 + 2*y + 2))/(y + 1)^3 = -2*(y + 1)/(y+1)^4 = -2/(y+1)^3

I made mistake on this point.

Thank you so much for you answer and help.