Find the slope of the tangent line to each curve at the given point. y = square root x + 5; (4, 3)

Question

I am having trouble with this Calculus Question.

Answer 1

To find the slope of the tangent line, we take the derivative of y.

 

y = √(x + 5)

 

 

We can write this square-root function using exponents.

 

y = (x + 5)^1/2

 

 

We use the chain rule to derive this function.

 

y' = (1/2)(x + 5)^-1/2

 

y' = 1 / (2√(x + 5))

 

 

Next, we plug in the x coordinate to evaluate y'.

 

y' = 1 / (2(√(4 + 5))

 

y' = 1 / (2√(9))

 

y' = 1 / (2 * 3)

 

y' = 1/6

 

The slope of the tangent line is 1/6.

Answer 2

y = √(x+5)

 

dy/dx = ½(x+5)^-½

 

slope of tangent line at the point (4, 3) = ½(4+5)^-½

 

                                                                                      = 1/6