Differentiation of (y^4)(x^2)=128, and then given dy/dx = 4 show c = -2d and find the value of c and d.
been stuck on this question for a while now, any help would be appreciated thanks!
I have moved your topic to the Calculus sub-forum, as it involves differentiation.
Can you show what you have tried so far? You want to implicitly differentiate with respect to $x$. This means you are going to need to apply the product, power and chain rules. Can you find \(\displaystyle \frac{dy}{dx}\)?
thanks! when i differentiated i got : (-3x^2)(y^2)/(2x^3)(y).
i also forgot to mention in the question that the point where dy/dx=3 is (a,b).
I feel im right so far but when it comes to working out the coordinates, i cant get an answer, thanks again!
i managed to cancel down to give 6a^3/-3a^2=b which then simplified to b=-2a,
however when i sub the b=-2a into the question i dont get an answer? does that mean a=0? or am i doing it all wrong? thanks
Can you do the substitution? You should find a non-zero value for $a$. I want to see your work when you make the substitution, to see how you are getting zero for $a$.
i got x=2! thanks! however in the question it says the point P lies on curve V and has coordinates (a,b). So i was trying to sub the b=-2a into the dy/dx equation
You obtained \(\displaystyle b=-2a\) from the equation:
\(\displaystyle \frac{dy}{dx}=-\frac{3y}{2x}\)
so substituting this result back into it will only result in an identity as I showed above.
You now have that \(\displaystyle a=2\), so what is $b$?
By the way, you will get much better help if you give the problems exactly as stated in their entirety up front. Giving the entire problem, along with the directions, and all work you have done is the best way to get prompt, pertinent help.
I'm not "fussing," but rather offering you helpful advice.