Hard Q to take a derivative of

In summary: Which is more or less equivalent to the previously given answers. I guess the point is that there are many ways to do the same thing. In summary, the conversation discusses different methods for taking the derivative of the given function, including using the quotient and product rules, partial fractions, logarithmic differentiation, and expanding the denominator. Each method may vary in elegance and preference, but all yield the same result.
  • #1
the4thcafeavenue
14
0
y=(x)/((x+2)(x+3)(x+4)).
how to do u take the derivative? HELP!
 
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  • #2
1) DON'T PANIC!

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.
 
  • #3
good work arildno
 
  • #4
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.
 
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  • #5
I recommend OP to pay close attention to Curious' excellent procedure.
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
arildno said:
I recommend OP to pay close attention to Curious' excellent procedure.
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.

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

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
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.
 
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  • #8
Shouldn't that be an (x + 4) in there? Bottom line, on the left side of the last = sign. It's probably just a typo.
 
  • #9
Yes it was.
 
  • #10
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 )
 
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Related to Hard Q to take a derivative of

1. How do I take the derivative of a function?

Taking the derivative of a function involves finding the slope of a curve at a specific point. This can be done by using the power rule, chain rule, product rule, or quotient rule depending on the complexity of the function.

2. What is the purpose of taking a derivative?

The derivative is used to calculate the rate of change of a function at a specific point, which can be useful in many fields such as physics, engineering, and economics.

3. Can any function be differentiated?

Not all functions can be differentiated, as some may not have a well-defined slope at certain points. For example, a function with a sharp corner or a discontinuity cannot be differentiated.

4. What is the difference between a derivative and an antiderivative?

A derivative is the instantaneous rate of change of a function, while an antiderivative is the inverse operation of differentiation, also known as integration.

5. How can I check if my calculated derivative is correct?

You can check your calculated derivative by using the derivative rules and comparing your result to the original function. Additionally, you can use graphing software to plot the original function and its derivative to visually verify the correctness of the derivative.

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