Equation of tangent line?

Question

Write the equation of the tangent line to

1.) x² +2x −5y³ = −2 at the point (1, 1).

√x+√y=9 at x=4

 

Answer 1

First, we differentiate x² + 2x - 5y³ = -2 on both sides of equation using chain rule.  This is method is known as implicit differentiation.  Keep in mind that the derivative is the slope of the tangent line

 

(x² + 2x - 5y³)' = (-2)'

 

2x + 2 - 15y²y' = 0

 

Solve for y'.

 

-15y²y' = -2x - 2

 

y' = (-2x - 2)/(-15y²)

 

Substitute the values of the point (1,1) into y'.

 

y' = [-2(1) - 2] / [-15(1)²]

y' = -4 / (-15)

y = 4/15

 

This will be the slope of the tangent line.

 

We want the equation of the tangent line in the form y=mx+b where m is slope and b is y-intercept.  Substitute the values of the point (1,1) to find b.

 

y = mx + b

1 = (4/15)(1) + b

1 = 4/15 + b

 

11/15 = b

 

Your equation of the tangent line is

 

y = (4/15)x + (11/15)

 

 

Now that you know how to find the equation of the tangent line, try out the second part of the problem.