# slope of line tanget to graph

• Dec 23rd 2007, 10:16 AM
DINOCALC09
slope 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, 10:20 AM
Jhevon
Quote:

Originally Posted by DINOCALC09
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.

how else would you do it?

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

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

differentiating implicitly we get:

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

now solve for dy/dx and plug in the required x- and y-values
• Dec 23rd 2007, 10:25 AM
DINOCALC09
Quote:

Originally Posted by Jhevon
how else would you do it?

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

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

differentiating implicitly we get:

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

now solve for dy/dx and plug in the required x- and y-values

dy/dx = -6x/(5/y)?
• Dec 23rd 2007, 10:28 AM
Jhevon
Quote:

Originally Posted by DINOCALC09
dy/dx = -6x/(5/y)?

no.

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

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

$\Rightarrow \frac {dy}{dx} = \frac {-6xy}5$
• Dec 23rd 2007, 10:32 AM
DINOCALC09
so as im understanding, the answer is (A) or -12/5
• Dec 23rd 2007, 10:54 AM
Jhevon
Quote:

Originally Posted by DINOCALC09
so as im understanding, the answer is (A) or -12/5

yes