Quiotient rule problem

[skipqtexas]

I'm having a problem of combining like terms and simplifying.

The problem is y=-3x+5/3x+4

When I do the problem according to formula v (x) x u (x)- u (x) x v (x) / (v (x) )^2 get -9x-12 - -9x+15

But when you combine like terms an simplify the final answer is -27/ 3x+4^2

How do you get that (simplifying like terms)?

 

[skipqtexas]

help anybody

 

[stapel]

I'm having a problem of combining like terms and simplifying.

The problem is y=-3x+5/3x+4

What you have posted means the following:

. . . . .\(\displaystyle y\, =\, -3x\, +\, \dfrac{5}{3x}\, +\, 4\)

I suspect, however, that you omitted the (necessary, in typed form) grouping symbols, so you meant something more like y = (-3x + 5)/(3x + 4):

. . . . .\(\displaystyle y\, =\, \dfrac{-3x\, +\, 5}{3x\, +\, 4}\)

Also, "the problem", as posted, has no instructions. What are you supposed to be doing with this? You mention "combining like terms and simplifying" (algebraic topics), but the "formula" you mention later seems more related perhaps to differentiation (a calculus topic)...?

 

[Jason76]

The quotient rule only deals with derivatives:

Quotient Rule: \(\displaystyle \dfrac{(g)(f') - (f)(g')}{g^{2}}\) given \(\displaystyle \dfrac{d}{dx}(\dfrac{f}{g})\)

\(\displaystyle y = \dfrac{3x + 5}{3x + 4}\)

\(\displaystyle y' = \dfrac{(3x + 4)(3) - (3x + 5)(3)}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{(9x + 12) - (9x + 15)}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{9x + 12 - 9x - 15}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{0 - (-3)}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{0 + 3}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{3}{(3x + 4)^{2}}\) Answer

 

[skipqtexas]

The quotient rule only deals with derivatives:

Quotient Rule: \(\displaystyle \dfrac{(g)(f') - (f)(g')}{g^{2}}\) given \(\displaystyle \dfrac{d}{dx}(\dfrac{f}{g})\)

\(\displaystyle y = \dfrac{3x + 5}{3x + 4}\)

\(\displaystyle y' = \dfrac{(3x + 4)(3) - (3x + 5)(3)}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{(9x + 12) - (9x + 15)}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{9x + 12 - 9x - 15}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{0 - (-3)}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{0 + 3}{(3x + 4)^{2}}\)

\(\displaystyle y' = \dfrac{3}{(3x + 4)^{2}}\) Answer

This problem is the quotient of a derivative. I'm doing Quotient problems on the differentiation course on my software.

It says find dy/dx given the following. y= u (x)/ v (x)

Dervative of Quotient

d/dx (u/v)= v du/dx - v du/dx / v^2

5x^2+3/ 4x^2 +2 = 40x^3+20 - 40x^3+24

Which becomes -4/(4x^2+2)^2 when simplified and cmbining like term. Somon please explain how answer is -4/4x^2+2

 

[JeffM]

This problem is the quotient of a derivative. I'm doing Quotient problems on the differentiation course on my software.

It says find dy/dx given the following. y= u (x)/ v (x)

Dervative of Quotient

d/dx (u/v)= v du/dx - v du/dx / v^2

5x^2+3/ 4x^2 +2 = 40x^3+20 - 40x^3+24

Which becomes -4/(4x^2+2)^2 when simplified and cmbining like term. Somon please explain how answer is -4/4x^2+2

Please be careful. You were told earlier to use grouping symbols; you do not. You mean the derivative of a quotient I suspect, not quotient of a derivative. And pay no attention to Jason; he is a student, not a tutor.

Is this the problem:

\(\displaystyle Find\ the\ derivative\ with\ respect\ to\ x\ of\ y = \dfrac{5x^2 + 3}{4x^2 + 2}?\)

If that is the problem you must show it as (5x^2 + 3) / (4x^2 + 2). The parentheses are important.

\(\displaystyle u = 5x^2 + 3 \implies \dfrac{du}{dx} = 10x.\)

\(\displaystyle v = 4x^2 + 2 \implies \dfrac{dv}{dx} = 8x.\)

\(\displaystyle y = \dfrac{5x^2 + 3}{4x^2 + 2} = \dfrac{u}{v} \implies \dfrac{dy}{dx} = \dfrac{\left(v * \dfrac{du}{dx}\right) - \left(u * \dfrac{dv}{dx}\right)}{v^2} \implies\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{\{(4x^2 + 2) * 10x\} - \{(5x^2 + 3) * 8x\}}{(4x^2 + 2)^2} = \dfrac{40x^3 + 20x - (40x^3 + 24x)}{(4x^2 + 2)^2} = \dfrac{40x^3 + 20x - 40x^3 - 24x}{(4x + 2)^2} = \dfrac{-4x}{(4x + 2)^2} \ne \dfrac{-4}{(4x + 2)^2}.\)

I do not get your answer.

 

[skipqtexas]

Please be careful. You were told earlier to use grouping symbols; you do not. You mean the derivative of a quotient I suspect, not quotient of a derivative. And pay no attention to Jason; he is a student, not a tutor.

Is this the problem:

\(\displaystyle Find\ the\ derivative\ with\ respect\ to\ x\ of\ y = \dfrac{5x^2 + 3}{4x^2 + 2}?\)

If that is the problem you must show it as (5x^2 + 3) / (4x^2 + 2). The parentheses are important.

\(\displaystyle u = 5x^2 + 3 \implies \dfrac{du}{dx} = 10x.\)

\(\displaystyle v = 4x^2 + 2 \implies \dfrac{dv}{dx} = 8x.\)

\(\displaystyle y = \dfrac{5x^2 + 3}{4x^2 + 2} = \dfrac{u}{v} \implies \dfrac{dy}{dx} = \dfrac{\left(v * \dfrac{du}{dx}\right) - \left(u * \dfrac{dv}{dx}\right)}{v^2} \implies\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{\{(4x^2 + 2) * 10x\} - \{(5x^2 + 3) * 8x\}}{(4x^2 + 2)^2} = \dfrac{40x^3 + 20x - (40x^3 + 24x)}{(4x^2 + 2)^2} = \dfrac{40x^3 + 20x - 40x^3 - 24x}{(4x + 2)^2} = \dfrac{-4x}{(4x + 2)^2} \ne \dfrac{-4}{(4x + 2)^2}.\)

I do not get your answer.

How do I write out my problem like shown above?

 

[stapel]

How do I write out my problem like shown above?

Use LaTeX ("LAY-teck") formatting. You can see the coding when you do "Reply with Quote", such as:

[tex]u = 5x^2 + 3 \implies \dfrac{du}{dx} = 10x.[/tex]

[tex]v = 4x^2 + 2 \implies \dfrac{dv}{dx} = 8x.[/tex]

[tex] y = \dfrac{5x^2 + 3}{4x^2 + 2} = \dfrac{u}{v} \implies  \dfrac{dy}{dx} 

     = \dfrac{\left(v * \dfrac{du}{dx}\right) - \left(u *  \dfrac{dv}{dx}\right)}{v^2} \implies[/tex]

[tex]\dfrac{dy}{dx} = \dfrac{\{(4x^2 + 2) * 10x\} - \{(5x^2 + 3) *  8x\}}{(4x^2 + 2)^2}

    = \dfrac{40x^3 + 20x - (40x^3 + 24x)}{(4x^2 + 2)^2}

    =  \dfrac{40x^3 + 20x - 40x^3 - 24x}{(4x + 2)^2}

    = \dfrac{-4x}{(4x + 2)^2}  \ne \dfrac{-4}{(4x + 2)^2}.[/tex]

The above is what displayed the nice formatting you quoted.

 

[JeffM]

How do I write out my problem like shown above?

Use LaTeX, but I do not recommend that students fuss with it. Just take care with grouping symbols and the like.

I prefer to use LaTeX in answering students because it is so clear and tracks what students see in their texts, but students should probably concentrate their attention on doing math rather than formatting it elegantly.