PLEASE HELP! (Waiting for an hour..._) Find an equation of the tangent line to the curve at the given point??
$\ \Large \frac{x^2}{16}-\frac{y^2}{9} =1$
$\ \Large \text{The point is: } (-5, \frac{9}{4}) $.
PLEASE HELP!

Question

PLEASE HELP! (Waiting for an hour..._) Find an equation of the tangent line to the curve at the given point?? 
$\ \Large \frac{x^2}{16}-\frac{y^2}{9} =1$
$\ \Large \text{The point is: } (-5, \frac{9}{4}) $.
PLEASE HELP!

anonymous · Answer

what math are you in?

anonymous · Answer

AP Calculus BC

anonymous · Answer

what's "BC"

anonymous · Answer

It's just a level of calculus.... I need to find the derivative of this equation...

anonymous · Answer

solve for y and take the derivative?

anonymous · Answer

This section involves implicit differentiation

anonymous · Answer

ohhh

anonymous · Answer

did you take the derivative?

anonymous · Answer

No I'm still stuck on this problem!!

anonymous · Answer

I don't know how to with those fractions

anonymous · Answer

the derivative of $\frac{x^2}{16}$ is $\frac{x}{8}$

anonymous · Answer

Really? Wouldn't the 16 become a zero?

anonymous · Answer

and the derivative with respect to $x$ of $-\frac{y^2}{9}$ is 
$$-\frac{2y}{9}y'$$

anonymous · Answer

no it is a constant, think 
$$\frac{x^2}{16}=\frac{1}{16}x^2$$

anonymous · Answer

Am I just using power rule here? I don't need to use quotient rule (that's what I was thinking...?)

anonymous · Answer

@satellite73 I've been getting mixed answers from people. When do I know when I take the derivative of y to have yy' versus just y'???