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

Question

anonymous · Answer

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

saifoo.khan · Answer

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

anonymous · Answer

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

anonymous · Answer

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

saifoo.khan · Answer

and?

saifoo.khan · Answer

wht about -5/4?

anonymous · Answer

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

saifoo.khan · Answer

can u please come back to ques..??

anonymous · Answer

sure link to it

saifoo.khan · Answer

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

anonymous · Answer

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)

anonymous · Answer

did you get this kehara? any question?

anonymous · Answer

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