X (dy/dx) = (y-1/y+1)-y

1 answer:

8 0

F' $( \frac{y-1}{y+1})-y$ =
f' $( \frac{y-1}{y+1})$ -f'(y) [/tex]=
$\frac{f'(y-1)(y+1)-f'(y+1)(y-1)}{(y+1)^2} -1$ =
$\frac{(f'(y)-f'(1))(y+1)-(f'(y)+f'(1))(y-1)}{(y+1)^2} -1$ =
$\frac{(1-0)(y+1)-(1-0)(y-1)}{(y+1)^2} -1$ =
$\frac{(1)(y+1)-(1)(y-1)}{(y+1)^2} -1$ =
$\frac{(y+1)-(y-1)}{(y+1)^2} -1$ =
$\frac{y+1-y+1}{(y+1)^2} -1$ =
$\frac{2}{(y+1)^2} -1$

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If f(x) = ln(2), then limx--->2 (f(2)-f(x))/x-2

Blizzard [7]

Answer:

as written, -2
with denominator parentheses, 0
with f(x)=ln(x) and denominator parentheses, -1/2

Step-by-step explanation:

The problem as stated asks for the limit as x approaches 2 of (0/x) -2.

As written, the limit is (0/2) -2 = -2.

Explanation: f(x) is a constant, so the numerator is 0. The ratio 0/x -2 is defined as -2 everywhere except x=0. So, the value at x=2 is 0/2 -2 = -2.

If you mean (f(2) -f(x))/(x -2), that limit is the limit of 0/(x-2) = 0 as x approaches 2.

Explanation: f(x) is a constant, so the numerator is 0. The ratio 0/(x-2) is zero everywhere except at x=2. The left limit and right limit are both 0 as x approaches 2. Since these limits agree, the limit is said to be 0.

If you mean f(x) = ln(x) and you want the limit of (f(2) -f(x))/(x -2), that value will be -1/2.

Explanation: The value of the ratio is 0/0 at x=2, so we can find the limit using L'Hôpital's rule. Differentiating numerator and denominator, we have ...

lim = (-1/x)/(1)

The value is -1/2 at x=2.