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3 years ago

8

The product of a mixed number between 4 and 5 and a fraction between 0 and 1 is always less then 4?

1 answer:

topjm [15]3 years ago

5 0

The answer to the problem is true

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There is an existing logarithmic property that states that $log \frac{M}{N}$ is equal to $log M-logN$ .

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3 years ago

What is the value of \dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) dx d ( 3x 2 −4 2x+3 )start fraction, d, divided by, d, x

stellarik [79]

Answer:

4.

Step-by-step explanation:

We are asked to find the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ .

First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:

$(\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$

$\frac{d}{dx}(\frac{2x+3}{3x^2-4})$

$\frac{\frac{d}{dx}(2x+3)*(3x^2-4)-(2x+3)*\frac{d}{dx}(3x^2-4)}{(3x^2-4)^2}$

$\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}$

$\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}$

$\frac{-6x^2-18x-8}{(3x^2-4)^2}$

Therefore, our required derivative is $\frac{-6x^2-18x-8}{(3x^2-4)^2}$ .

Now, we will substitute $x=-1$ in our derivative to find the required value as:

$\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}$

$\frac{-6(1)+18-8}{(3(1)-4)^2}$

$\frac{-6+18-8}{(3-4)^2}$

$\frac{4}{(-1)^2}$

$\frac{4}{1}$

$4$

Therefore, the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ is 4.

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3 years ago