How would I find the slope of a tangent line?

Answer 1

The previous guys answer was excellent but I thought I'd break it down in seemingly smaller steps.

1) Find the derivative of the equation.
2) Solve that derivative equation at the point that the tangent line meets the curve.

Let's take the simple equation of f(x) = x^2 and the slope of the tangent at point (3,2).
1) The derivative of f(x) = x^2 is 2x.
2) Solve f(x) = 2x at (3,2) we get 6.

The slope of the tangent line of f(x) = x^2 at (3,2) is 6.

Answer 2

The slope of a tangent line can be find by first finding out the equation of the curve(as a function of x) & then differentiating it with respect to x. The coordinates of the point where the line is tangent to the curve are then put into the differentiated equation & the value obtained is the slope.

Eg.
The equation of the curve is y^2 = 16x & we have to find the slope of the tangent at (1,4).
Then,
Now, differentiating it with respect to x, we get
2y (dy/dx) = 16

dy/dx = 16/2y

x=1 & y=4.
Putting in equation,
dy/dx = 16/2*4 =2

So, the slope is 2.

Answer 3

perhaps pythagoras would help...
(half-circle over a base line gives the touching point with a 90 degrees' angle - the rest should be possible to solve)

Answer 4

Rise over run or Y/X over a discrete distance.