Lcd Chain
Lcd Chain
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Chain rule & LCD?
Hello, I would like it if someone could go through the follwing problems throughly.
y = x(1-x)^1/2
y' = 1 (1-x^2)^1/2 + x [1/2 (1 -x ^2)^-1/2 (2x)]
How can I use the LCD to solve this? My teacher requires me to solve by using the LCD; she said she will take points off if we do not do it like this on the test.
Here is another one:
y= x(x^2 + 1)^-1/2
y'= 1(x^2 +1)^1/2 + x[-x (x^2 +1) ^-3/2]
Again, how can I use the LCD to solve this?
I would think that I find common factors first, multiply it out second, and collect like terms last. However, each time I have done this I get the wrong answer.
First of all, there is a typo somewhere in the presentation of the first problem
If this is the original function,
y = x(1-x)^1/2 THEN
why is there a (1 - x^2) in the derivative?
*****************************************************
On your second function,
y= x(x^2 + 1)^-1/2
I think that there is an error in getting its derivative as presented above.
I am getting this ...
(dy/dx) = 1[x^2 + 1]^-1/2 + x[-1/2((x^2 + 1)^-3/2)(2x)
(dy/dx) = (1/(x^2 + 1)^1/2) - (x^2/(x^2 + 1)^3/2)
Your LCD is (x^2 + 1)^3/2 and so
(dy/dx) = (x^2 + 1 - x^2) / (x^2 + 1)^3/2
Simplifying the numerator,
(dy/dx) = 1 / (x^2 + 1)^3/2


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