Well I got this question:

Differentiate from first principals (i did it) $\displaystyle f=x^3 $ $\displaystyle f'(x) = 3x^2 $

Find the equation of the normal to the curve at point where x = 3.

So first I found y. $\displaystyle y=(3)^3 y=27 $ That means we got $\displaystyle (3,27) $

Then I subbed 3 into $\displaystyle 3x^2 $ to get the gradient of the tangent. Answer is $\displaystyle 27 $

Since the normal and tangent are perpendicular the formula is $\displaystyle m_{2} = \frac {-1}{m_{1}} $ So that makes $\displaystyle \frac {-1}{27} $

Then I subbed it into $\displaystyle y-y_{1} = m(x-x_{1}) $

$\displaystyle y-27=\frac{-1}{27}(x-3) $

** <=== you used the wrong sign in the bracket**

Finally I get

$\displaystyle -27y+726=x+3 $ but

**for some reason** **<=== maybe they believe that 27² = 729**
the answers in the book says it's different

Did I do anything wrong?

Also, I need help with another question:

If $\displaystyle y=x^3-3x^2 $ find the derivative and find the values of y when $\displaystyle \frac{dy}{dx} = 0 $

How would I go about starting this one?