Respuesta :
Answer:
4.
Step-by-step explanation:
We are asked to find the value of expression [tex]\frac{d}{dx}(\frac{2x+3}{3x^2-4})[/tex] at [tex]x=-1[/tex].
First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:
[tex](\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}[/tex]
[tex]\frac{d}{dx}(\frac{2x+3}{3x^2-4})[/tex]
[tex]\frac{\frac{d}{dx}(2x+3)*(3x^2-4)-(2x+3)*\frac{d}{dx}(3x^2-4)}{(3x^2-4)^2}[/tex]
[tex]\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}[/tex]
[tex]\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}[/tex]
[tex]\frac{-6x^2-18x-8}{(3x^2-4)^2}[/tex]
Therefore, our required derivative is [tex]\frac{-6x^2-18x-8}{(3x^2-4)^2}[/tex].
Now, we will substitute [tex]x=-1[/tex] in our derivative to find the required value as:
[tex]\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}[/tex]
[tex]\frac{-6(1)+18-8}{(3(1)-4)^2}[/tex]
[tex]\frac{-6+18-8}{(3-4)^2}[/tex]
[tex]\frac{4}{(-1)^2}[/tex]
[tex]\frac{4}{1}[/tex]
[tex]4[/tex]
Therefore, the value of expression [tex]\frac{d}{dx}(\frac{2x+3}{3x^2-4})[/tex] at [tex]x=-1[/tex] is 4.
