Question

find an equation for the line tangent to the curve at the given point
y= (x^4 +2)/(x^2)
at x=-1

Answer 1

When you don't have a 0/0 you can most of the time just simply use the substitution method.

Answer 2

which gives you lim= 3

Answer 3

whoops wrong question lol give me a second

Answer 4

Do you know how to use the quotient rule?

Answer 5

Yes that is what we just learned but I am a little confused on how to apply it in this scenario

Answer 6

Do you know how to find the d/dx?

Answer 7

would the d/dx be 2x-(4/x^3) ?

Answer 8

Yeap

Answer 9

so then do i need to just plug in -1 and then that would be the slope and then put the slope into either point slope or slope intercept form to get the equation of the tangent?

Answer 10

yes

Answer 11

That should give you your answer

Answer 12

great!!! thank you so much..
one more question if you could help me out and do not mind..
how do i find the d/dx of
y= 2(sqrt(x)) - (1/x)

Answer 13

\[y=2\sqrt{x}-1/x\] This?

Answer 14

yes please

Answer 15

oh actually its\[y = 2\sqrt{x} - 1/\sqrt{x}\]

Answer 16

ahh ok

Answer 17

The easiest way I find is to convert the sqrt into fractions.

Answer 18

ok once i have it in a fraction how do i proceed?

Answer 19

You differentiate the sum term by erm and factor out the constants

Answer 20

So 2(d/dx(sqrt(x)))-d/dx (1/Sqrt(x))

Answer 21

So you break them up into pieces.

Answer 22

ok. so then the first term, would then end up as?
2(1/2 x^-1/2) which simplifies to x^-1/2

Answer 23

well 1/Sqrt(x)= -1/2x^(3/2)

Answer 24

yes your right on the first term

Answer 25

so then would it simplify down to
y1= x^(-1/2) - 1/2x^(3/2)

Answer 26

The cleanest way to write it would be \[\frac{ 2x+1 }{ 2x^\frac{ 3 }{ 2 } }\]

Answer 27

where did you get the 2x+1

Answer 28

\[\frac{ 1 }{ 2x^ \frac{ 3 }{ 2 } }+ \frac{ 1 }{ \sqrt{x} }\]

Answer 29

thank you so much!!!

Answer 30

np