SOLUTION: Hello Everyone Having Trouble with this. Evaluate: dy=dx at x = 0 where (i) y = (x^2 - 1)/(x^2 + 2x + 1) (ii) y = (e^x + 1)/(1 + e^-x) Thank You

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Hello Everyone Having Trouble with this. Evaluate: dy=dx at x = 0 where (i) y = (x^2 - 1)/(x^2 + 2x + 1) (ii) y = (e^x + 1)/(1 + e^-x) Thank You      Log On


   

Question 375478: Hello Everyone
Having Trouble with this.

Evaluate:
dy=dx at x = 0 where
(i) y = (x^2 - 1)/(x^2 + 2x + 1)
(ii) y = (e^x + 1)/(1 + e^-x)
Thank You
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
+y+=+%28x%5E2+-+1%29%2F%28x%5E2+%2B+2x+%2B+1%29+
y=%28x%5E2-1%29%2F%28x%2B1%29%5E2
Use the quotient rule,
+dy%2Fdx+=+%28%28x%2B1%29%5E2%282x%29-%28x%5E2-1%29%282%28x%2B1%29%29%29%2F%28x%2B1%29%5E4
dy%2Fdx=%282x%28x%2B1%29-2%28x%5E2-1%29%29%2F%28x%2B1%29%5E3
dy%2Fdx=%282x%5E2%2B2x-2x%5E2%2B2%29%29%2F%28x%2B1%29%5E3
dy%2Fdx=%282%28x%2B1%29%29%2F%28x%2B1%29%5E3
dy%2Fdx=2%2F%28x%2B1%29%5E2
When x=0,
dy%2Fdx=2%2F%280%2B1%29%5E2
highlight%28dy%2Fdx=2%29
.
.
.
y=%28e%5Ex+%2B+1%29%2F%281+%2B+e%5E%28-x%29%29

dy%2Fdx=%28%28e%5E%28x%29%2B1%2B1%2Be%5E%28-x%29%29%29%2F%281%2Be%5E%28-x%29%29%5E2
dy%2Fdx=%28%28e%5E%28x%29%2Be%5E%28-x%29%2B2%29%29%2F%281%2Be%5E%28-x%29%29%5E2
dy%2Fdx=%28%28e%5E%28x%29%2Be%5E%28-x%29%2B2%29%29%2F%281%2Be%5E%28-x%29%29%5E2
dy%2Fdx=%28%28e%5E%28x%29%2Be%5E%28-x%29%2B2%29%29%2F%28e%5E%28-2x%29%2B2e%5E%28-x%29%2B1%29
dy%2Fdx=e%5E%28x%29
When x=0,
dy%2Fdx=e%5E%280%29
highlight%28dy%2Fdx=1%29