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Answer :

Supernovago

Answer:

[tex]\displaystyle \frac{d}{dx} \Big [ \frac{3}{x^2} \Big ]=-\frac{6}{x^3}[/tex]

Step-by-step explanation:

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

Power Rule: [tex]\displaystyle \frac{d}{dx}[ x^n] = nx^n^-^1[/tex]  

Rewrite [tex]\displaystyle \frac{3}{x^2}[/tex] as [tex]\displaystyle 3\cdot \frac{1}{x^2}[/tex] which is equivalent to [tex]3\cdot x^-^2[/tex].

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 [tex]\displaystyle -\frac{6}{x^3}[/tex].

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

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

  • [tex]\displaystyle \frac{(x^2)(0)-(3)(2x)}{(x^2)^2}[/tex]

Simplify.

  • [tex]\displaystyle \frac{-6x}{x^4}[/tex]

Using exponent rules, we can rewrite this as:

  • [tex]-6x^1^-^4 = -6x^-^3[/tex]

This can be rewritten as [tex]\displaystyle -\frac{6}{x^3}[/tex], which is the same as what we got using the product rule.

Jacob193go

Answer:

[tex]\displaystyle \frac{d}{d x}\left[\frac{3}{x^2}\right] = -\frac{6}{x^{3}}[/tex].

Step-by-step explanation:

Let [tex]f(x)[/tex] and [tex]g(x)[/tex] denote two functions of [tex]x[/tex]. Assume that [tex]g(x) \ne 0[/tex]. The quotient rule states that:

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

In this question:

  • The numerator of the fraction is [tex]f(x) = 3[/tex] (a constant function.)
  • The denominator of the fraction is [tex]g(x) = x^{2}[/tex].

Find [tex]{f}^{\prime}(x)[/tex] and [tex]{g}^{\prime}(x)[/tex].

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

For [tex]{g}^{\prime}(x)[/tex], consider the power rule:

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

Apply this rule to find [tex]{g}^{\prime}(x)[/tex].

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

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

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

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

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