Given the equation:
[tex]w(x)=\frac{15}{x^2+1}[/tex]Let's find the equation of the line tangent to the graph of the given function at the point (2, 3)
Let's find the first derivative and evaluate for the values at x = 2, y = 3.
We have:
Differentiate w(x) using Constant Multiple rule
[tex]\begin{gathered} 15\frac{d}{dx}\lbrack\frac{1}{x^2+1}\rbrack \\ \\ =15\frac{d}{dx}\lbrack(x^2+1)^{-1}\rbrack \end{gathered}[/tex]Next is to differentiate using chain rule:
Rewrite u for (x²+1)
[tex]\begin{gathered} 15(\frac{d}{du}\lbrack u^{-1}\rbrack\frac{d}{dx}\lbrack x^2+1\rbrack) \\ \\ =15(u^{-2}\frac{d}{dx}\lbrack x^2+1\rbrack) \\ \\ =15(-(x^2+1)^{-2}\frac{d}{dx}\lbrack x^2+1\rbrack) \\ \\ =-15(x^2+1)^{-2}(2x+0) \\ \\ =-30(x^2+1)^{-2}x \\ \\ =\frac{-30x}{(x^2+1)^2} \end{gathered}[/tex]Now, evaluate the derivative when x = 2.
Substitute 2 for x and evaluate:
[tex]\begin{gathered} \frac{-30(2)}{(2^2+1)^2} \\ \\ =\frac{-60}{(4+1)^2} \\ \\ =\frac{-60}{(5)^2} \\ \\ =\frac{-60}{25} \\ \\ =-2.4 \end{gathered}[/tex]Thus, the slope of the tangent line is -2.4.
Apply the point-slope form of a linear equation:
y - y1 = m(x - x1)
Where m is the slope.
Input the values of (2, 3) for x1 and y1. Then substitute -2.4 for m:
[tex]y-3=-2.4(x-2)[/tex]Let's solve the euqtion for y.
Apply distributive property to right side of the equation:
[tex]\begin{gathered} y-3=-2.4x-2.4(-2) \\ \\ y-3=-2.4x+4.8 \end{gathered}[/tex]Add 3 to both sides of the equation:
[tex]\begin{gathered} y-3+3=-2.4x+4.8+3 \\ \\ y=-2.4x+7.8 \end{gathered}[/tex]Therefore, the equation of the line tangent to the graph of the given function is:
y = -2.4x + 7.8
ANSWER:
[tex]y=-2.4x+7.8[/tex]