calculus
posted by Pam .
Please help, my live tutor and I could spent 2 hours trying to solve these two:
find an equation for tangent line y=18^2@ (4,129)
find the derivative of the function:we both kept getting the same answer, but Mathlab kept saying that it was wrong: 4lnl3xl/2+3x
we did the quotient rule, and the chain rule...the derivative and our sign would be wrong...

I cannot make sense out of your equation, check your typing.

4 ln l3xl
2+3x
4ln absolute value of 3x over 2+3x
y= 18x^2
(4,129)
the first set is in fraction form.
the 3x is absolute.
the 8x^2 is 8x squared. 
for the derivative using the quotient rule I got (8  12x + 12x(ln(3x))) / (x(2+3x)^2)
for the tangent question
dy/dx = 16x
at the point the slope is 16(4) = 64
so using y = mx + b for the tangent equation
129 = 64(4) + b
b = 127
so the tangent equation is y = 64x + 127 
Thanks. One small error? I can see what we did wrong..we did not do the
129= part.....guess we were tired, or tried to do it in our heads.