Thread: differentiate the following:

hello, I am hoping someone can help me understand a step.

Q1.

y = In [ (x+2) /
sqrt (x-1)]

y = In (x+2) - 1/2 In (x-1)

dy/ dx = 1/x+2 - 1/2 (1/x-1)

Now I don't understand where, on top, x-4. How do you get that??

Answer is dy/dx = x-4 / 2(x+2)(x-1)

Another example.

Q2.
y = In [ (x+1) /
sqrt (x-1)]

answer is = x-3 / 2(x^2 -1) . how come?

Many thanks in advance!!

Originally Posted by Jon123

hello, I am hoping someone can help me understand a step.

Q1.

y = In [ (x+2) /
sqrt (x-1)]

y = In (x+2) - 1/2 In (x-1)

dy/ dx = 1/x+2 - 1/2 (1/x-1)

Now I don't understand where, on top, x-4. How do you get that??

Answer is dy/dx = x-4 / 2(x+2)(x-1)


What about common denominator??

$\displaystyle \frac{1}{x+2}-\frac{1}{2(x-1)}=\frac{2(x-1)-(x+2)}{2(x-1)(x-2)}=\frac{x-4}{2(x-1)(x+2)}$


Another example.

Q2.
y = In [ (x+1) /
sqrt (x-1)]

answer is = x-3 / 2(x^2 -1) . how come?


Same thing as above. This is basic algebra...

Tonio


Many thanks in advance!!

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