hey guys...i just had 2 questions that i really needed help with, i have no clue how to even start these question, any help would be greatly appreciated, thx
Question 1
Find the equation of the tangent line to the curve y = x^2 ln x at the point (1,0).
the answer should be in the form y =...
ok. i think i got it
y = (cosx)^sinx
logy = log(cosx^sinx)
logy=sinx(logcosx)
(dy/dx)(1/(yln10)) = sinx(1/cosx(ln10) + log(cosx^cosx)
(dy/dx)=[(tanx/ln10) + (log(cosx^cosx))](yln10)
jus let me know if I am on the right track. thanks for your help guys.
Using log differentiation, find dy/dx, in terms of x for the following:
y = (cosx)^sinx
any help wud be appreciated, I am unsure of how to start this question, thanks in advance
that looks about right, i matched up the data given and ur graph...and seems kinda accurate...one thing is that...ur f(-2) doesn't seem to be a point of inflection in ur graph its just going up... without changing direction or anything...so i think u might want to fix that