The slope of the line tangent to the graph of 3x^2 + 5ln(y) = 12 at (2,1) is...

a) -12/5

b) 12/5

c) 5/12

d) 12

e) -7

**Im not particularly differentiating y in respect to x and could use some help.

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- Dec 23rd 2007, 09:16 AMDINOCALC09slope of line tanget to graph
The slope of the line tangent to the graph of 3x^2 + 5ln(y) = 12 at (2,1) is...

a) -12/5

b) 12/5

c) 5/12

d) 12

e) -7

**Im not particularly differentiating y in respect to x and could use some help. - Dec 23rd 2007, 09:20 AMJhevon
how else would you do it?

i'd differentiate implicitly here and then plug in x = 2, y = 1 into the formula i get.

$\displaystyle 3x^2 + 5 \ln y = 12$

differentiating implicitly we get:

$\displaystyle 6x + \frac 5y~\frac {dy}{dx} = 0$

now solve for dy/dx and plug in the required x- and y-values - Dec 23rd 2007, 09:25 AMDINOCALC09
- Dec 23rd 2007, 09:28 AMJhevon
no.

$\displaystyle 6x + \frac 5y~\frac {dy}{dx} = 0$ ........move the 6x to the other side by subtracting it from both sides

$\displaystyle \Rightarrow \frac 5y ~\frac {dy}{dx} = -6x$ ........multiply both sides by $\displaystyle \frac y5$

$\displaystyle \Rightarrow \frac {dy}{dx} = \frac {-6xy}5$ - Dec 23rd 2007, 09:32 AMDINOCALC09
so as im understanding, the answer is (A) or -12/5

- Dec 23rd 2007, 09:54 AMJhevon