# Thread: slope of line tanget to graph

1. ## 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.

2. 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.

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

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

3. 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.

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

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

dy/dx = -6x/(5/y)?

4. Originally Posted by DINOCALC09
dy/dx = -6x/(5/y)?
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 $\displaystyle \frac y5$

$\displaystyle \Rightarrow \frac {dy}{dx} = \frac {-6xy}5$

5. so as im understanding, the answer is (A) or -12/5

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