# B Some doubts about these differentiation steps

1. May 2, 2017

### awholenumber

First of all there is an equation
Then there is the derivative
Then there is a point slope formula to find the equation of the tangent line
Point slope formula to obtain the tangent line .
y=3a2(x-a)+a3
Then Plug in the x coordinate into the derivative to get the slope

f'(1) = 3(1)2
f'(1) = 3

What this means is that for any value of x=a, the instantaneous slope of f at (a,a3) is 3a2.

Here , i don't really understand some change of terms from f(x) to y , a ... etc

2. May 2, 2017

### PeroK

When you draw a graph (using x and y axes), you are actually doing a graph of:

$y = f(x)$

And, in this case:

$y = x^3$

In one sense, $y$ and $f(x)$ are the same thing. And you can write the derivative in various forms $\frac{dy}{dx} = y'(x) = f'(x)$.

If you want the derivative at a point $x = a$, you can write: $f'(a) = y'(a) = \frac{dy}{dx}(a) = \frac{dy}{dx}|_{a} = \frac{dy}{dx}|_{x=a}$

You may see all these different notations. But, they all mean the same thing.

3. May 2, 2017

### Mastermind01

Do you know what a derivative is (physical significance)?

We also have the point slope form from coordinate geometry $y-y' = m(x-x')$ which gives the equation of the line if the slope and one point $(x',y')$ are known.

4. May 2, 2017

### awholenumber

Thanks for the reply ,

Actually i have many doubts , not sure where to start .

Lets start with the point itself , which is an x,y coordinate when the graph is x along the horizontal axis and y along the vertical axis . I hope this part is alright .

Anyway

We usually write a function like

f(x)=x3
y = f(x)
y = x3

Between two points on a curve there is a slope, or rate of change, that we can calculate as

$\frac{\Delta y}{\Delta x}$

With calculus and derivatives we are trying to find the instantaneous slope at one point. This is done by taking two points and shrinking them closer and closer and closer together, until they approach the same point. The goal is to find a rate that captures the slope of a function at a particular point

Point slope formula to find the equation of the tangent line

y-y1 = m (x - x1 )

So these equations are same as

f(x) = x3
f(x) = 3x2(x-x)+x3
x=0

??

Last edited: May 2, 2017
5. May 2, 2017

### PeroK

The slope of the curve (derivative) at a given point is a number. You can then imagine a straight line through that point with the same slope. THis is called the tangent line. That line has a straight line equation. For a curve like $x^3$ there is a different tangent line at every point. For example, at the point $(1, 1)$ the slope of the curve is $3$. The tangent line to the curve at this point is, therefore:

$y-1 = 3(x-1)$

We can also draw this line on the graph. But, if you have any two curves or lines on a graph, you have to be careful which one you are talking about.

In general, you can find the tangent line at any point on the curve $(a, a^3)$ in the same way.

And, you can find the tangent to any curve at any point in the same way.

$y - f(a) = f'(a)(x-a)$

In my first post I said that "in a sense" $y$ and $f(x)$ are the same thing, but if you have more than one curve, you need to show which one you are talking about - and $y$ can mean different things in different contexts.

6. May 2, 2017

### awholenumber

Thanks for helping ...

I am a bit confused about the coordinates , is this an y , x coordinate or an x , y coordinate ?

7. May 2, 2017

### PeroK

It's always $(x, y)$ by convention.

8. May 2, 2017

### awholenumber

Thanks a lot PeroK for all the replies .
If i ask more silly doubts this thread might turn into a mess . :)

I need to re read this thread a couple of times to completely understand it .