Question

Find dy over dxA) 2x+1 over x power 4 + 2

Answer 1

Explanation:

Given y defined below:

$y=\frac{2x+1}{x^4+2}$

To find the derivative of y, we use the quotient rule.

$\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

In the function:

$\begin{gathered} u=2x+1\implies\frac{du}{dx}=2 \\ v=x^4+2\implies\frac{dv}{dx}=4x^3 \end{gathered}$

Substitute these values into the formula:

$\frac{d}{dx}(\frac{u}{v})=\frac{2(x^4+2)-4x^3(2x+1)}{(x^4+2)^2}$

We then simplify:

$\begin{gathered} \frac{dy}{dx}=\frac{2x^4+4-8x^4-4x^3}{(x^4+2)^2} \\ =\frac{2x^4-8x^4-4+4x^3}{(x^4+2)^2} \\ =\frac{-6x^4+4x^3-4}{(x^4+2)^2} \\ =\frac{-2(3x^4+2x^3-2)}{(x^4+2)^2} \end{gathered}$

Answer:

The derivative dy/dx of the given function is:

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