Question

What is the slope of the line tangent to the graph of y = (9x^2)/(x+2) at x = 1 ?

Answer 1

Using derivatives, it is found that the slope of the line tangent to the graph of y at x = 1 is of 5.

The equation is:

$y(x) = \frac{9x^2}{x + 2}$

The slope of the line tangent to the graph of y at x = 1 is $y^{\prime}(1)$.

Applying the quotient rule, we have that:

$y^{\prime}(x) = \frac{[9x^2]^{\prime}(x + 2) - [x + 2]^{\prime}(9x^2)}{(x + 2)^2}$

$y^{\prime}(x) = \frac{18x(x + 2) - 9x^2}{(x + 2)^2}$

$y^{\prime}(x) = \frac{9x^2 + 36x}{(x + 2)^2}$

Then, when x = 1:

$y^{\prime}(1) = \frac{9(1)^2 + 36(1)}{(1 + 2)^2} = \frac{45}{9} = 5$

The slope of the line tangent to the graph of y at x = 1 is of 5.

You can learn more about the use of derivatives to find the slope of a tangent line at brainly.com/question/10580177