Question

Dy/dx f(x)= 2 /(3x+1)

Answer 1

The derivative of
`\(f(x) = (2)/(3x+1)\) is \((-6)/((3x+1)^2)\)` obtained using the quotient rule, where
`\(u(x) = 2\)` and
`\(v(x) = 3x+1\).`

To find the derivative
`\( (dy)/(dx) \)` of the given function
`\( f(x) = (2)/(3x+1) \),` you can use the quotient rule. The quotient rule states that if you have a function in the form
`\( (u(x))/(v(x)) \),` then the derivative is given by:

`\[ (d)/(dx)\left((u(x))/(v(x))\right) = (u'(x)v(x) - u(x)v'(x))/((v(x))^2) \]`

Let's apply this rule to your function:

`\[ u(x) = 2 \]`

`\[ v(x) = 3x + 1 \]`

Now, find the derivatives:

`\[ u'(x) = 0 \]` (since the derivative of a constant is zero)

`\[ v'(x) = 3 \]` (the derivative of
`\(3x + 1\)` with respect to
`\(x\)` is 3)

Now, apply the quotient rule:

`\[ (dy)/(dx) = (u'(x)v(x) - u(x)v'(x))/((v(x))^2) \]`

`\[ (dy)/(dx) = (0 \cdot (3x + 1) - 2 \cdot 3)/((3x + 1)^2) \]`

Simplify:

`\[ (dy)/(dx) = (-6)/((3x + 1)^2) \]`

So, the derivative of the given function
`\( f(x) = (2)/(3x+1) \)` with respect to
`\(x\)` is
`\( (-6)/((3x + 1)^2) \).`