Question

Use the quotient rule to find the derrivatives of 3/x^2

With full process ​

Answer 1

`\displaystyle (d)/(d x)\left[(3)/(x^2)\right] = -(6)/(x^(3))`.

Explanation:

Let
`f(x)` and
`g(x)` denote two functions of
`x`. Assume that
`g(x) \\e 0`. The quotient rule states that:

`\displaystyle (d)/(d x)\left[(f(x))/(g(x))\right] = \frac{{f}^\prime(x) \cdot g(x) - f(x) \cdot g^(\prime)(x)}{{(g(x))}^2}`.

In this question:

• The numerator of the fraction is
  `f(x) = 3` (a constant function.)
• The denominator of the fraction is
  `g(x) = x^(2)`.

Find
`{f}^(\prime)(x)` and
`{g}^(\prime)(x)`.

Notice that the value of
`f(x)` is constantly
`3` regardless of the value of
`x`. By the constant rule,
`{f}^(\prime)(x) = 0`.

For
`{g}^(\prime)(x)`, consider the power rule:

if
`m` represents a rational number (such as
`2`,) then by the power rule,
`\displaystyle (d)/(d x)\left[{x}^(m)\right] = m\, {x}^(m-1)`.

Apply this rule to find
`{g}^(\prime)(x)`.

`\begin{aligned}{g}^(\prime)(x) &= (d)/(d x) \left[x^(2)\right] \\ &= 2\, x^(2 - 1) \\ &= 2\, x\end{aligned}`.

Substitute
`f(x) = 3`,
`{f}^(\prime)(x) = 0`,
`g(x) = x^(2)`, and
`{g}^(\prime)(x) = 2\, x` into the quotient rule expression to find
`\displaystyle (d)/(d x)\left[\frac{3}{{x}^(2)}\right]`:

`\begin{aligned}&\; (d)/(d x)\left[\frac{3}{{x}^(2)}\right] \quad \genfrac{}{}{0em}{}{\leftarrow f(x) = 3}{\leftarrow g(x) = {x}^(2)} \\ =&\; (d)/(d x)\left[(f(x))/(g(x))\right]\\ =&\; \frac{{f}^\prime(x) \cdot g(x) - f(x) \cdot g^(\prime)(x)}{{(g(x))}^2} \\ =& \; \frac{0 \, x^2 - 3\, (2\, x)}{\left({x}^(2)\right)} \\ =&\; -(6)/(x^(3))\end{aligned}`.

Therefore,
`\displaystyle (d)/(d x)\left[\frac{3}{{x}^(2)}\right] = -(6)/(x^(3))`.

Answer 2

`\displaystyle (d)/(dx) \Big [ (3)/(x^2) \Big ]=-(6)/(x^3)`

Explanation:

We can use either the Power Rule or the Quotient Rule to find the derivative of 3/x².

Power Rule:
`\displaystyle (d)/(dx)[ x^n] = nx^n^-^1`

Rewrite
`\displaystyle (3)/(x^2)` as
`\displaystyle 3\cdot (1)/(x^2)` which is equivalent to
`3\cdot x^-^2`.

Now we can apply the power rule to 3x⁻². Subtract 1 from the exponent and multiply the coefficient by -2.

• 3(-2)x⁻³ = -6x⁻³

This can be rewritten as
`\displaystyle -(6)/(x^3)`.

Quotient Rule:
`\displaystyle (d)/(dx) \Big [ (f(x))/(g(x)) \Big ]=(g(x)f'(x)-f(x)g'(x))/([g(x)]^2)`

Substituting 3 for f(x) and x² for g(x), we get:

• 
  `\displaystyle ((x^2)(0)-(3)(2x))/((x^2)^2)`

Simplify.

• 
  `\displaystyle (-6x)/(x^4)`

Using exponent rules, we can rewrite this as:

• 
  `-6x^1^-^4 = -6x^-^3`

This can be rewritten as
`\displaystyle -(6)/(x^3)`, which is the same as what we got using the product rule.