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

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2 Answers

4 votes

Answer:


\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)).

User Jama
by
6.3k points
4 votes

Answer:


\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.

User Karen Tracey
by
6.5k points
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