derivatives

[maeveoneill]

okay these are pretty basic questions. just wondering if someone could check these over for me. there are two equations which i have derived to find the prime and double prime and also the intercepts.

1.
y=(x³ +16)/x
y'= (2x³-16)/x²
y''= (2x^4 +32x)/x^4

2.
y= (x²-4)/(x²-1)
y' = 6x/(x²-1)²
y'' = im stuck on this one as well.


for equation 1. the intercepts are (2,0)(-2,0)(0,4)
for equation 2. no intercepts exist

 

[galactus]

For f'' of the 2nd problem:

Try using the product rule instead of the quotient rule. I always thought it was a little less cumbersome.

\(\displaystyle (6x)(x^{2}-1)^{-2}\)

First times derivative of second plus second times derivative of first:

\(\displaystyle (6x)(2x)(-2(x^{2}-1)^{-3})+(x^{2}-1)^{-2}(6)\)

\(\displaystyle (12x^{2})(-2(x^{2}-1)^{-3})+6(x^{2}-1)^{-2}\)

\(\displaystyle (-24x^{2})(x^{2}-1)^{-3}+6(x^{2}-1)^{-2}\)

\(\displaystyle \frac{-24x^{2}}{(x^{2}-1)^{3}}+\frac{6}{(x^{2}-1)^{2}}\)

\(\displaystyle \frac{-24x^{2}+6(x^{2}-1)}{(x^{2}-1)^{3}\)

\(\displaystyle \frac{-18x^{2}-6}{(x^{2}-1)^{3}\)

\(\displaystyle \frac{-6(3x^{2}+1)}{(x^{2}-1)^{3}\)

 

[stapel]

For y" in (2), just apply the Quotient Rule again.

How did you obtain your intercepts for (1)? Or are you finding the intercepts of something other than y?

Eliz.

 

[maeveoneill]

For y" in (2), just apply the Quotient Rule again.

How did you obtain your intercepts for (1)? Or are you finding the intercepts of something other than y?

Eliz.

when x=0 and y=0.. are the rest of the derivatives correct?

 

[stapel]

when x=0 and y=0.

Does this refer to my question about the intercepts...? If so, then please reply showing the steps you did, and specifying which of y, y', and y" is the function for which you're finding the intercepts.

For instance, the function y in (1) isn't even defined for x = 0, so the point (0, 4) can't possibly be an intercept of its graph.

Eliz.