View Full Version : math problem

lizzpalmer

06-30-2011, 06:32 PM

Here's the lastest one lol

g(t) = t^3 + t - 2

(2t - 1)^5

this is what I think I should do:

3t^2 + 1 * (2t - 1)^-5

then

3t^2 + 1 -10t + 5

so

3t^2 -10t + 6

but that could be completely wrong lol

lizzpalmer

06-30-2011, 06:33 PM

I am trying to find the derivative of the function :)

mmm4444bot

06-30-2011, 07:05 PM

You could rewrite the ratio as a product, and then use the Product Rule for derivatives.

Or, you could leave it as given, and use the Quotient Rule for derivatives.

Does your list of formulas show a "Product Rule" ?

Here is what the function looks like, written as a product instead of a ratio:

g(t) = (t^3 + t - 2)(2t - 1)^{-5}

You must also apply the Chain Rule, whichever method you use.

Do you understand what the Chain Rule is?

lizzpalmer

06-30-2011, 07:38 PM

I have a video on the chain rule lol let me see if I can find it.

http://www.khanacademy.org/video/the-ch ... t=Calculus (http://www.khanacademy.org/video/the-chain-rule?playlist=Calculus)

mmm4444bot

06-30-2011, 07:57 PM

Do you understand the Chain Rule?

lizzpalmer

06-30-2011, 08:11 PM

I think I am starting to get it. I will try and post :)

lizzpalmer

06-30-2011, 08:21 PM

g ' (t) = (30x + 10) (2t-1)^-6

does that look right?

lizzpalmer

06-30-2011, 08:24 PM

that should be (30t + 10)(2t-1) sorry

mmm4444bot

07-01-2011, 02:36 AM

No — your results for g`(t) are not correct.

The examples of the Chain Rule in that Khan-Academy video use functions that are single polynomial-powers. In other words, there is only one polynomial raised to a power.

Your exercise is a product of polynomial-powers, and there is another rule for finding the derivative of a product of powers. It's called the Product Rule.

EGs:

The expression (2x + 3)^3 is a single polynomial raised to a power.

The expression (t^3 + t - 2)^(1) * (2t - 1)^(-5) is a product of polynomials; each polynomial is raised to its own power.

NOTE: Do not write the exponent 1; I typed that only to show more clearly the product of two powers.

You need to use both the Product Rule and the Chain Rule, when taking the derivative of g(t).

g(t) = (t^3 + t - 2)(2t - 1)^(-5)

You could follow the Khan example to find the derivative of the (2t - 1)^(-5) part, but that derivative is only one part of the calculations to find the first derivative of g(t). The rest of it comes from the Product Rule.

In English, the Product Rule says something like, "the derivative of a product is the first times the derivative of the second PLUS the second times the derivative of the first".

(t^3 + t - 2) is the "first" and (2t - 1)^(-5) is the "second"

I am not sure what materials you have to study.

I want to ask you: did your on-line course provide you the link to that video as a lesson, or did you find it on your own?

At this point, I must also ask whether or not your on-line course gave you any lessons on the Product Rule.

If you look at the calculus index at Khan Academy, there is a video on the Product Rule.

Cheers ~ Mark

lizzpalmer

07-03-2011, 10:21 AM

The instructor gave the links to the videos.

I will re-review this. I'm sort of lost as what to do next but I will look at it again.

lizzpalmer

07-03-2011, 10:39 AM

g(t) = (t^3 + t - 2) (2t - 1) ^-5

g ' (t) = (3t + 1) 2 * 5 (2t - 1)^-6 + (t^3 + t - 2) * (-10t^-6)

=(30t + 10)(2t-1)^-6 + (t^3 + t - 2) * 60t^-7

Would that be right? Or what am I doing wrong?

JeffM

07-03-2011, 02:32 PM

g(t) = (t^3 + t - 2) (2t - 1) ^-5

g ' (t) = (3t + 1) 2 * 5 (2t - 1)^-6 + (t^3 + t - 2) * (-10t^-6)

=(30t + 10)(2t-1)^-6 + (t^3 + t - 2) * 60t^-7

Would that be right? Or what am I doing wrong?

When you are starting out, it is much easier to use substitutions although it is longer. Once this is all familiar, you can skip the substitutions.

Let u = t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2.

du/dt = 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1. Polynomial rule.

Let v = 2t - 1

dv/dt = 2. Polynomial rule.

Let w = v[sup:8b0f5wmp]-5[/sup:8b0f5wmp] = (2t - 1)[sup:8b0f5wmp]-5[/sup:8b0f5wmp].

dw/dv = -5v[sup:8b0f5wmp]-6[/sup:8b0f5wmp] = -5(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]. Polynomial rule.

dw/dt = (dw/dv) * (dv/dt) = -10(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]. Chain rule.

Let F(t) = x = uw.

F'(t) = dx/dt = u(dw/dt) + w(du/dt). Product rule.

F'(t) = (t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2)(-10(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]) + (2t - 1)[sup:8b0f5wmp]-5[/sup:8b0f5wmp](3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1) =

(-10(t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2) + (2t - 1)(3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1)) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp] =

(-10t[sup:8b0f5wmp]3[/sup:8b0f5wmp] - 10t + 20 + 6t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + 2t - 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] - 1) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp]=

(-4t[sup:8b0f5wmp]3[/sup:8b0f5wmp] - 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] - 8t + 19) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp] =

-(4t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 8t - 19) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp].

The least error-prone way to find the derivative of a complex expression is to use substitutions and then use the basic rules of differentiation such as the product rule, the quotient rule, and the chain rule. As I said above, with more experience, you can skip the substitutions.

lizzpalmer

07-03-2011, 09:43 PM

That's great - I haven't seen it like that before. It makes a lot more sense! Thank you

mmm4444bot

07-04-2011, 10:46 AM

Having read your previous comments about the course and your instructor, I looked at Southern New Hampshire University's web site. The 500-level on-line course that you're taking could very well be designed to weed out people whom the university does not want in their Master's program. Generally, students spend more than one year studying algebra, precalculus, calculus, and linear algebra; the university has bundled these four courses into this single on-line course, so it must be review. It could also be a test to determine whether you are capable of completing a Master's degree on-line. This would explain why you're not getting a lot of direct instruction or support from the university (eg: referring you to the Khan Academy).

I am glad that Jeff's explanation makes sense to you. I wish you continued success.

In this thread, function g is a ratio of functions. I suggested rewriting it as a product and using the Product Rule, as one method, but I also mentioned the Quotient Rule. The Quotient Rule provides the derivative without rewriting the ratio as a product.

Quotient Rule:

If f(x) = g(x)/h(x), where functions g and h are differentiable and h(x) is not zero, then the derivative of f(x) is:

f`(x) = [h(x) * g'(x) - g(x) * h'(x)]/h(x)^2

In English, we could describe this rule as "The bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared".

Using the Quotient Rule on (t^3 + t - 2)/(2t - 1)^5 yields:

\frac{(2t - 1)^{5} \cdot (3t^{2} + 1) - (t^3 + t - 2) \cdot 5(2t - 1)^{4}(2)}{(2t - 1)^{10}}

This result is the derivative of your g(t), and it simplifies to the same result coming out of the Product Rule in Jeff's post.

Cheers 8-)

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