I am new in differential equation. I just want to solve the following problem,

Can any one give me the step wise solution for this question???

Determine the equation of the tangent line for the function

f(x) = x² + 1 at point (3,10).

Find the slope of the function by differentiation
f '(x) = 2x

Plug in the certain point's values Since this function does not have y we don't plug in y yet
f '(3) = 6 {6 is now the slope of the point 3,10}

Plug both slope and point values into a linear equation
(y - y₁) = m(x - x₁) {this is the linear equation}
(y - 10) = 6(x - 3)  {Which can be simplified as below}
y = 6x -8

Just as we can find the slope and equation of a tangent line for a function, we can also do the same for a normal line. However, the normal line has two differences from the tangent line.

1. The slope of a normal line is perpendicular to the slope of the tangent line. Or in other words, the negative inverse of the tangent line.

2. The normal line is only defined if x does not = 0.

As a result, to find the slope and equation of the normal line, follow the steps above and convert the slope of the tangent line to the slope of the normal line.