Please help, my live tutor and I could spent 2 hours trying to solve these two:

Question

find an equation for tangent line y=-1-8^2@ (-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...

Answer 1

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

Answer 2

-4 ln l3xl
2+3x
-4ln absolute value of 3x over 2+3x

y= -1-8x^2
(-4,-129)
the first set is in fraction form.
the 3x is absolute.
the 8x^2 is 8x squared.

Answer 3

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

Answer 4

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.