# find the slope pf the tangent to the curve at the given point:?

1. y=7-x^2=4x^3 at (2,1)

2.y^2-x^2-1 at (3,2)

Please..help me..i need it badly for my removal exam.....hope someone can help me with this..tnx n advance

### 2 Answers

- 8 years agoFavorite Answer
1.

did you mean : y=7-x^2+4x^3 then

dy/dx=-2x+12x^2

x=2 --> dy/dx=2x+12x^2 = 2*2+12*2^2 = 52

x=1 --> dy/dx=2x+12x^2 = 2*1+12*1^2 = 14

2. y^2=1+x^2 -->y=±√(1+x^2)

dy/dx=±x/√(1+x^2)

x=3 --> dy/dx=±x/√(1+x^2) = ±3/√(10)

x=2 --> dy/dx=±x/√(1+x^2) = ±2/√(5)

- K DLv 48 years ago
Whenever you have to find the tangent line to the curve at a given point, all you do is:

1. take the derivative of the function.

2. plug in the x-value that is given to you.

Now understand that the first two steps will give you the slope of the tangent line at that x-value.

3. The tangent line of any function is a line. You need two things to make a line: a point and a slope. You have both. Use the point given to you and the slope and plug it into the point slope formula.

y - y1 = m ( x - x1)

That will be the equation of the tangent line at that point.