Question

answer is given, explain the full process including final steps

Answer 1

Given the curve:

$y=\frac{18}{x^2+2}$

You need to find the equation for the tangent to that curve at the point:

$(1,6)$

Then, you need to follow these steps:

1. Derivate the function given in the exercise using these Derivative Rules:

$\frac{d}{dx}(\frac{1}{u(x)})=-\frac{u^{\prime}(x)}{u(x)^2}$$\frac{d}{dx}(x^n)=nx^{n-1}$$\frac{d}{dx}(k)=0$

Where "k" is a constant.

Then, you get:

$=18\cdot\frac{(x^2+2)^{\prime}}{(x^2+2)^2}$$=-18\cdot\frac{(x^2+2)^{\prime}}{(x^2+2)^2}$$=-18\cdot\frac{2x}{(x^2+2)^2}$$y^{\prime}=-\frac{36x}{(x^2+2)^2}$

2. Substitute this value of the x-coordinate of the given point into the derivated function and then evaluate, in order to find the slope of the line:

$m=-\frac{36(1)}{(1^2+2)^2}=-\frac{36}{9}=-4$

4. The Point-Slope Form of the equation of a line is:

$y-y_1=m(x-x_1)$

Where "m" is the slope of the line and this point is on the line:

$(x_1,y_1)$

In this case:

$\begin{gathered} m=-4 \\ x_1=1 \\ y_1=6 \end{gathered}$

Therefore, you can substitute values:

$y-6=-4(x-1)$

5. Convert the equation from Point-Slope Form to Slope-Intercept Form.

The equation of a line in Slope-Intercept Form is:

$y=mx+b$

Where "m" is the slope and "b" is the y-intercept.

Then, by solving for "y", you get:

$y-6=(-4)(x)-(-4)(1)$

$y-6=-4x+4$$y=-4x+4+6$$y=-4x+10$

Hence, the answer is:

1. Derivate the function.

2. Find the slope of the tangent line by substituting the x-coordinate of the given point into the function derivated.

3. Write the equation of the tangent line in Point-Slope Form using the given point and the slope.

4. Rewrite the equation in Slope-Intercept Form by solving for "y".

Equation for the tangent of the curve:

$y=-4x+10$