Friday

April 18, 2014

April 18, 2014

Posted by **Pam** on Friday, April 25, 2008 at 7:59am.

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...

- calculus -
**Reiny**, Friday, April 25, 2008 at 8:21amI cannot make sense out of your equation, check your typing.

- calculus -
**Pam**, Friday, April 25, 2008 at 11:39am-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.

- calculus -
- calculus -
**Reiny**, Friday, April 25, 2008 at 12:33pmfor 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

- calculus -
**Pam**, Friday, April 25, 2008 at 12:45pmThanks. 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.

- calculus -

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