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(a) If
F(x)=5x//(1+x^(2)), find
F^(')(2) and use it to find an equation of the tangent line to the curve
y=5x//(1+x^(2)) at the point
(2,2).
(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

$31$ . (a) If $F(x)=5 x /\left(1+x^{2}\right)$ , find $F^{\prime}(2)$ and use it to find an equation of the tangent line to the curve $y=5 x /\left(1+x^{2}\right)$ at the point $(2,2)$ . $\newline$ (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

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Find derivatives of inverse trigonometric functions

Full solution

Q.

31

. (a) If

F(x)=5 x /\left(1+x^{2}\right)

, find

F^{\prime}(2)

and use it to find an equation of the tangent line to the curve

y=5 x /\left(1+x^{2}\right)

at the point

(2,2)

.

\newline

(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

Apply Quotient Rule: To find the derivative $F'(x)$ of the function $F(x) = \frac{5x}{1+x^2}$ , we will use the quotient rule, which states that if you have a function that is the quotient of two functions, $\frac{u(x)}{v(x)}$ , then its derivative is given by $\frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$ . Let $u(x) = 5x$ and $v(x) = 1 + x^2$ . Then $u'(x) = 5$ and $v'(x) = 2x$ . $F'(x) = \frac{(v(x)u'(x) - u(x)v'(x))}{(v(x))^2}$ $F'(x) = \frac{((1 + x^2)(5) - (5x)(2x))}{(1 + x^2)^2}$ $F(x) = \frac{5x}{1+x^2}$ $0$ $F(x) = \frac{5x}{1+x^2}$ $1$
Calculate Derivative: Now we need to evaluate $F'(x)$ at $x = 2$ to find the slope of the tangent line at the point $(2,2)$ . $\newline$ $F'(2) = \frac{5 - 5(2)^2}{(1 + (2)^2)^2}$ $\newline$ $F'(2) = \frac{5 - 20}{(1 + 4)^2}$ $\newline$ $F'(2) = \frac{-15}{25}$ $\newline$ $F'(2) = \frac{-3}{5}$
Evaluate at $x=2$ : With the slope of the tangent line at the point $(2,2)$ being $-\frac{3}{5}$ , we can use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is the point on the line. $\newline$ Using the point $(2,2)$ and the slope $-\frac{3}{5}$ , we get: $\newline$ $y - 2 = (-\frac{3}{5})(x - 2)$
Use Point-Slope Form: To write the equation of the tangent line in slope-intercept form $y = mx + b$ , we will solve for $y$ .
$y = \frac{-3}{5}(x - 2) + 2$
$y = \frac{-3}{5}x + \frac{6}{5} + 2$
$y = \frac{-3}{5}x + \frac{6}{5} + \frac{10}{5}$
$y = \frac{-3}{5}x + \frac{16}{5}$
This is the equation of the tangent line.
Convert to Slope-Intercept Form: For part (b), graphing the curve and the tangent line would require plotting the function $y = \frac{5x}{1+x^2}$ and the line $y = \frac{-3}{5}x + \frac{16}{5}$ on the same set of axes. This step is more about using graphing software or graphing by hand, and it's not a mathematical calculation step, so we will not perform it here.

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