Question

Find an equation of the tangent line 1/2 {x log(x^2), {-1, 0}}

Answer 1

do you know how to find slope of tangent line ?

Answer 2

I need to take the derivative using the product rule but having trouble

Answer 3

I get 1/2(ln(x^2)) + 1/2(x)1/x^2(2x) Don't know if that is right,

Answer 4

you'll need chain rule to differentiate log (x^2)
yes! thats correct :)

Answer 5

ok cool. The other person that tried to help me said it was wrong. lol

Answer 6

So from that I would need to somehow get the slope and then use y - y1 = m(x - x1)

Answer 7

simplify it .

Answer 8

well it would be 1/2(ln(x^2) + (x)1/x^2(2x)

Answer 9

and then perhaps 1/2ln(x^2) + 1/x^2(2x^2)

Answer 10

for 2nd term.
1/2 and 2 gets cancelled
2 'x's in numerator gets cancelled with x^2 in denominator.
can you see it ??
so, 2nd term is just =1

Answer 11

so does that change to 1/2 ln(x^2) + 2 ?

Answer 12

And then how do I get the slope out of that?

Answer 13

wait, would I plug in the -1? into x?

Answer 14

If I did that I would get 1/2ln(-1^2) + 2

Answer 15

and then 1/2 ln(1)+2

Answer 16

hmm, do you have any suggestions?

Answer 17

So can someone tell me if this is right? (y-0) = (1/2ln(1)+2)(x-(-1).

Answer 18

yeah and I missed ) at the end

Answer 19

hey Tech-Hunter do you understand the problem?

Answer 20

sorry, i was disconnected.
1/2(x)1/x^2(2x) simplifies to +1 and not +2.
so,slope =1/2 ln(1)+1 is correct.
so, equation will be
y = (1/2ln(1)+1)(x+1).

Answer 21

1/2(x)1/x^2(2x) simplifies to +1
did u understand this ?

Answer 22

\[\dfrac{1}{2}x\dfrac{1}{x^2}2x=\dfrac{2}{2}\dfrac{x.x}{x^2}=1\]