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y=(x)/((x+2)(x+3)(x+4)).

how to do u take the derivative??? HELP!!!!!!

how to do u take the derivative??? HELP!!!!!!

- Thread starter the4thcafeavenue
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- #1

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y=(x)/((x+2)(x+3)(x+4)).

how to do u take the derivative??? HELP!!!!!!

how to do u take the derivative??? HELP!!!!!!

- #2

arildno

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2) Define [tex]u(x)=(x+2)(x+3)(x+4)[/tex]

3) Use this so that we have: [tex]y(x)=\frac{x}{u(x)}[/tex]

By the rule for differentiating a ratio between 2 functions, we have:

[tex]\frac{dy}{dx}=\frac{1*u(x)-x\frac{du}{dx}}{u^{2}}=\frac{1}{u}-\frac{x}{u^{2}}\frac{du}{dx}[/tex]

4) Evaluate [tex]\frac{du}{dx}[/tex] by use of the product rule.

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good work arildno

- #4

Curious3141

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I'd personally prefer to use a partial fractions method.

Let [tex]y = \frac{a}{x+2} + \frac{b}{x+3} + \frac{c}{x+4}[/tex]

Multiply throughout by [itex](x+2)(x+3)(x+4)[/itex], you get

[tex]x = a(x+3)(x+4) + b(x+2)(x+4) + c(x+2)(x+3)[/tex]

Substitute in turn [itex]x = -2, -3, -4[/itex] and almost immediately you get

[tex]a = -1, b = 3, c = -2[/tex]

giving [tex]y = \frac{-1}{x+2} + \frac{3}{x+3} + \frac{-2}{x+4}[/tex]

then

[tex]y' = \frac{1}{(x+2)^2} - \frac{3}{(x+3)^2} + \frac{2}{(x+4)^2}

[/tex]

and I would be happy to leave it in that form.

EDIT : Doing it this way is optional for the derivative (arildno's method is fine here), but partial fractions are essential for integrating the function.

Let [tex]y = \frac{a}{x+2} + \frac{b}{x+3} + \frac{c}{x+4}[/tex]

Multiply throughout by [itex](x+2)(x+3)(x+4)[/itex], you get

[tex]x = a(x+3)(x+4) + b(x+2)(x+4) + c(x+2)(x+3)[/tex]

Substitute in turn [itex]x = -2, -3, -4[/itex] and almost immediately you get

[tex]a = -1, b = 3, c = -2[/tex]

giving [tex]y = \frac{-1}{x+2} + \frac{3}{x+3} + \frac{-2}{x+4}[/tex]

then

[tex]y' = \frac{1}{(x+2)^2} - \frac{3}{(x+3)^2} + \frac{2}{(x+4)^2}

[/tex]

and I would be happy to leave it in that form.

EDIT : Doing it this way is optional for the derivative (arildno's method is fine here), but partial fractions are essential for integrating the function.

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arildno

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While it may seem very cumbersome and unnecessary to do the partial fractions decomposition, learning that technique through following Curious' example is only to your later benefit.

Also, it brings to light that there are always more than one single correct way of doing maths.

Good work, Curious.

- #6

Curious3141

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Thank you, arildno, for your very kind words.arildno said:

While it may seem very cumbersome and unnecessary to do the partial fractions decomposition, learning that technique through following Curious' example is only to your later benefit.

Also, it brings to light that there are always more than one single correct way of doing maths.

Good work, Curious.

It's completely true, you can never learn too many techniques in Math. Each new technique you learn is another invaluable tool you have in tackling a new problem.

- #7

Galileo

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In spirit of aildno's wise words, here's another method that gives you the derivative quickly, but doesn't look quite as beautiful as Curious' method.

It's logarithmic differentiation, take the logarithm of both sides:

[tex]\ln y = \ln x - \ln (x+2) -\ln (x+3) -\ln (x+4)[/tex]

then differentiate and use the chain rule:

[tex]\frac{dy}{dx}=y\left(\frac{1}{x}-\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}\right)=\frac{x}{(x+2)(x+3)(x+4)}\left(\frac{1}{x}-\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}\right)[/tex]

I`m not very fond of this method, but it serves its purpose. Points of worry might be the points where y is nonpositive, so you cannot take the logarithm. The method works just as well with the logarithm of the absolute value, so you can ignore that 'detail' when working out the problem. the only point remaining is where y=0, but that works out ok as well.

It's logarithmic differentiation, take the logarithm of both sides:

[tex]\ln y = \ln x - \ln (x+2) -\ln (x+3) -\ln (x+4)[/tex]

then differentiate and use the chain rule:

[tex]\frac{dy}{dx}=y\left(\frac{1}{x}-\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}\right)=\frac{x}{(x+2)(x+3)(x+4)}\left(\frac{1}{x}-\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}\right)[/tex]

I`m not very fond of this method, but it serves its purpose. Points of worry might be the points where y is nonpositive, so you cannot take the logarithm. The method works just as well with the logarithm of the absolute value, so you can ignore that 'detail' when working out the problem. the only point remaining is where y=0, but that works out ok as well.

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- #9

Galileo

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Yes it was.

- #10

uart

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It's interesting to see all the different methods and different (but equivalent) solutions that such a simple question can receive.

If I had received this question I'd probably have used the following simple (though none too elegant) method.

**y=x / ( (x+2)(x+3)(x+4) )**

Since the denominator is very easy to expand I'd write,

**y= x / (x^3 + 9x^2 + 26x + 24)** or **y = x (x^3 + 9x^2 + 26x + 24)^(-1)**

And then use either the quotient rule or the product and chain rule as per personal preference to eventually yield,

**dy/dx = (-2x^3 - 9x^2 + 24) / ( (x+2)^2 (x+3)^2 (x+4)^2 )**

If I had received this question I'd probably have used the following simple (though none too elegant) method.

Since the denominator is very easy to expand I'd write,

And then use either the quotient rule or the product and chain rule as per personal preference to eventually yield,

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