Question

Find an equation of the tangent line to the curve at the point (1, 1).

Answer 1

\[y=\ln(xe ^{x ^{5}})\]

Answer 2

satelite? i want the answer in compunt form.. with infinty and all.

Answer 3

i would first rewrite this as
\[\ln(x)+\ln(e^{x^5})=\ln(x)+x^5\] and go form there

Answer 4

@saifoo i guess you should have
\[(-\frac{1}{2},\infty)\]

Answer 5

and?

Answer 6

wht about -5/4?

Answer 7

that is it. compound means "both" as far as i understand it.

Answer 8

can u please come back to ques..??

Answer 9

sure link to it

Answer 10

http://openstudy.com/groups/mathematics/updates/4e1668930b8bc22757461111

Answer 11

y=ln(xe^x^5)
y=lnx+ln e^x^5
y=lnx+x^5 ln e
y=lnx+x^5
dy/dx = i/x +5x^4
at point(1,1) slope m=1/1 + 5(1)^4=6
y-y1=m(x-xi)
y-1=6(x-1)
y=6x-6+1
y+6x-5 ans......eq of the tangent line at pt(1,1)

Answer 12

did you get this kehara? any question?

Answer 13

oops sorry
it is
y=6x-5 ans...lol