Question

How do you find the equation of the tangent line to the curve at the given point. `y=5x−2sqrtx`, `(1,3)`?.

Answer 1

`y=4x-1`

Explanation:

Let `y=5x - 2sqrtx`
`=5x-2x^(1/2)`
`:. dy/dx = 5-x^(-1/2)`
`=5-sqrtx/x`

Let `x=1` to solve for the gradient at this point:

`:. dy/dx = 4`

Solve for the tangent equation by `y-y_1 = m(x-x_1)`

`:. y -3 =4(x-1)`
`:. y = 4x - 1`