Hi,

Can someone please tell me how to differentiate this particular formula with step by step direction?

A= 2n * (1/2) * (150/n) * [ (150/n) / tan (180/n) ]

Thanks in advance. As soon as possible if you can :)

Oops forgot to simplify tehe

A= (300n/2n) * [ (150/n) / tan (180/n) ]

You should be simplifying your equation first, further than what you've done:

\begin{eqnarray}A &=& \cancel{2n} \times \frac{1}{\cancel{2}} \times \frac{150}{\cancel{n}} \times \[\frac{\frac{150}{n}}{tan(\frac{180}{n})}\] \\ &=& 150 [ \frac{150 tan(\frac{180}{n})}{n}] \\ &=& \frac{22500tan(\frac{180}{n})}{n}\end{eqnarray}

Now use the quotient rule to differentiate:

If y = \frac uv

then.

\frac{dy}{dx} = \frac{v.\frac{du}{dx} - u.\frac{dv}{dx}}{v^2}

Hey thanks for the response. One question though, why did you put the tan (180/n) up? I don't that part.

I prefer dealing with simple fractions as far as possible. You can use the quotient rule, but to get less confused, putting it up is better.

If you're asking me if that's possible, then yes;

\frac{2}{\(\frac12\)} = 2 \div \frac 12 = 2 \times \frac 21 = \frac{2 \times 2}{1}

Be careful.

\frac{\frac{22500}{n}}{tan(\frac{180}{n})}=\frac{2 2500}{ntan(\frac{180}{n})}\neq \frac{22500tan(\frac{180}{n})}{n}

=\frac{22500cot(\frac{180}{n})}{n}

If that were \frac{1}{tan(\frac{180}{n})} in the denominator, then that would be OK.

Oh, sorry... I my head was not completely on that thread... :( Thanks galactus for correcting me.