Answer:
[tex]\frac{3}{16}[/tex]
Step-by-step explanation:
Hello, I think I can help you with this
let's remember
[tex]\frac{1}{x^{n} } =x^{-n} \\and\\\frac{dy}{dx} x^{n}=nx^{n-1}[/tex]
Let
[tex]f(x)=-\frac{3}{x} \\f(x)=y\\y=-\frac{3}{x} \\[/tex]
Step 1
derive
derivative of f(x)=derivative of y=y'
[tex]y=-\frac{3}{x}\\y=-3*x^{-1}\\ y'=-3(-1x^{-1-1})\\ y'=-3(-1x^{-2})\\y'=3x^{-2}\\ y'=\frac{3}{x^{2}}[/tex]
Step 2
fin y'(-4)
Put the value of -4 in the equation
[tex]y'=\frac{3}{x^{2}}\\ y'(-4)=\frac{3}{(-4)^{2}}\\y'(-4)=\frac{3}{16}[/tex]
Have a great day
