Question

If F(x) = 5x/(1 + x2), find F '(2) and use it to find an equation of the tangent line to the curve y = 5x/(1 + x2)at the point (2, 2).

Answer 1

Answer:

3x+5y-16=0

Step-by-step explanation:

See it in the pic.

Answer 2

Equation of tangent line to the curve is

$y=-\frac{3}{5}x+\frac{16}{5}$

Tangent line to the curve.

Equation of tangent line is in the form of y=mx+b

where m is the slope and b is the y intercept

Derivative :

Derivative of a given function is the slope. To find slope at the given point , we find out f'(2)

we are given with function

$f(x)= \frac{5x}{1+x^2} \\Apply \; quotient \; rule \\5\frac{\frac{d}{dx}\left(x\right)\left(1+x^2\right)-\frac{d}{dx}\left(1+x^2\right)x}{\left(1+x^2\right)^2}\\5\cdot \frac{1\cdot \left(1+x^2\right)-2xx}{\left(1+x^2\right)^2}\\\frac{5\left(1-x^2\right)}{\left(1+x^2\right)^2}$

Now we find f'(2) by replacing 2 for x

$f'(x)=\frac{5\left(1-x^2\right)}{\left(1+x^2\right)^2}\\f'(2)=\frac{5\left(1-2^2\right)}{\left(1+2^2\right)^2}\\f'(2)=\frac{-15}{25} \\f'(2)=-\frac{3}{5}$

Now we use the given point and the  slope to get the equation of tangent line

$m=-\frac{3}{5} and (2,2)\\y-y_1=m(x-x_1)\\y-2=-\frac{3}{5}(x-2)\\y-2=-\frac{3}{5}x+\frac{6}{5}\\y=-\frac{3}{5}x+\frac{6}{5}+2\\y=-\frac{3}{5}x+\frac{16}{5}$

Equation of tangent line to the curve is

$y=-\frac{3}{5}x+\frac{16}{5}$

Learn more about the tangent line here:

brainly.com/question/6617153