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
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 $\displaystyle \frac y5$
$\displaystyle \Rightarrow \frac {dy}{dx} = \frac {-6xy}5$