y=tan^-1(5ax/a^2-6x^2)show that dy/dx=3a/a^2+9x^2+2a/a^2+4x^2

Question

chaise · Answer

If you write this in equation editor I might be able to help.

anonymous · Answer

can you write the equation more simple???

jhannybean · Answer

@hartnn

jhannybean · Answer

quotient rule? I have no idea why it's dy/dx....

hartnn · Answer

my first thoughts will be tan y =(5ax/(a^2-6x^2)) ....find sec^2 y from here,
then differentiate both sides....
i'll try on paper first whether this works....

jhannybean · Answer

I have been...i don't understand where dy comes into the picture...lol

hartnn · Answer

when you differentiate y, with respect to x, you get dy/dx

jhannybean · Answer

so you're solving for y '...

hartnn · Answer

since, they have written dy/dx and not, $\partial y/\partial x$, we need to assume that a is constant here..

hartnn · Answer

yes, we are proving that by solving for y'

jhannybean · Answer

$$\large \tan^{-1}(\frac{5ax}{a^2-6x^2})$$Show that $$\large \frac{dy}{dx}=\frac{3a}{a^2}+9x^2+\frac{2a}{a^2}+4x^2$$

jhannybean · Answer

Omg this is too lengthy.

nincompoop · Answer

you signed up for it :P

nincompoop · Answer

I think it is supposed to be
[3a/(a^2+9x^2)]+[2a/(a^2+4x^2)]

nincompoop · Answer

I might be wrong since I don't know this stuff :X