1. ## A few questions

Firstly, I'm not sure which section this goes in (so please don't infract me for it )

a) Differentiate from first principles $\displaystyle y=x^2+2$

I get $\displaystyle 2x$

b) Find the gradient of the secant through the points. A(1,3) and B(3,11)

What is a secant?

c) Find the equation of the tangent to the curve at the point A

How would I go about doing this?

2. Hi jgv115,

Originally Posted by jgv115
Firstly, I'm not sure which section this goes in (so please don't infract me for it )
Calculus subforum is right.

Originally Posted by jgv115
a) Differentiate from first principles $\displaystyle y=x^2+2$

I get $\displaystyle 2x$
This is the right answer, and I believe "from first principles" means use the definition of derivative as limit, and don't just use the shortcut of taking the exponent and moving it in front and subtracting one from the exponent.

Originally Posted by jgv115
b) Find the gradient of the secant through the points. A(1,3) and B(3,11)

What is a secant?
This is just the line connecting the two points on a curve. MathWorld.

Originally Posted by jgv115
c) Find the equation of the tangent to the curve at the point A

How would I go about doing this?
You can get the slope from your answer to part (a), now you can use point-slope equation.

3. a) Assuming that you have evaluated $\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ and haven't simply used the rule, then you are correct.

b) A secant is a line that cuts through a curve in two or more points. So in other words, work out the equation of the line between $\displaystyle A(1, 3)$ and $\displaystyle B(3,11)$.

c) You know point $\displaystyle A(1,3)$ lies on the tangent line. So you can substitute $\displaystyle (x,y) = (1,3)$ into $\displaystyle y = mx+ c$. You can also evaluate $\displaystyle m$, since the derivative tells you the gradient of the tangent line at any point (so evaluate it at $\displaystyle x = 1$). Then you can solve for $\displaystyle c$.

4. Originally Posted by Prove It

c) You know point $\displaystyle A(1,3)$ lies on the tangent line. So you can substitute $\displaystyle (x,y) = (1,3)$ into $\displaystyle y = mx+ c$. You can also evaluate $\displaystyle m$, since the derivative tells you the gradient of the tangent line at any point (so evaluate it at $\displaystyle x = 1$). Then you can solve for $\displaystyle c$.
Sorry if I am missing something simple

I get the first bit $\displaystyle (x,y) = (1,3)$ but how do I evaluate m?

5. Originally Posted by jgv115
Sorry if I am missing something simple

I get the first bit $\displaystyle (x,y) = (1,3)$ but how do I evaluate m?
Didn't you go over the geometric interpretation of derivative?

6. Evaluate $\displaystyle \frac{dy}{dx}$ at the point $\displaystyle x = 1$.

7. Alright so $\displaystyle \frac{dy}{dx} = 2x$ Sub 1 in is $\displaystyle 2$

So $\displaystyle (y-y_{1}) = m(x-x_{1})$

$\displaystyle y=2x+1$

Is this correct?

8. Originally Posted by jgv115
Alright so $\displaystyle \frac{dy}{dx} = 2x$ Sub 1 in is $\displaystyle 2$

So $\displaystyle (y-y_{1}) = m(x-x_{1})$

$\displaystyle y=2x+1$

Is this correct?
Yes.

9. Originally Posted by jgv115
Alright so $\displaystyle \frac{dy}{dx} = 2x$ Sub 1 in is $\displaystyle 2$

So $\displaystyle (y-y_{1}) = m(x-x_{1})$

$\displaystyle y=2x+1$

Is this correct?
Yes