Abinashregmi01234idn
Answered

Get clear, concise, and accurate answers to your questions on IDNLearn.com. Our platform offers comprehensive and accurate responses to help you make informed decisions on any topic.

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

With full process ​

Sagot :

Supernovaidn

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.

Jacob193idn

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].

Your participation is crucial to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.