Derivative Calculations Question

[Silvanoshei]

Finding the first and 2nd derivatives here.

How would I start finding the 1st derivative of r = 1 / 3s² - 5 / 2s ?

I tried putting them in neg. exponent form, like r = 3s-² ... but that doesn't work for 5/2s? Do I need to use the Product Rule? Quotient is for division problems right? This is just subtraction?

 

[JeffM]

Finding the first and 2nd derivatives here.

How would I start finding the 1st derivative of r = 1 / (3s²) - 5 / (2s) ? Please use brackets or parentheses to ensure proper order of operations. What you wrote is not, I strongly suspect, what you mean.

I tried putting them in neg. exponent form, Perfectly good plan to use the power rule although you can also use the quotient rule like r = 3s-² ... but that doesn't work for 5/2s? Of course it works for 5/(s2) = (5/2)s^(-2). Do I need to use the Product Rule? If you take this approach you will need the Power Rule, the Product Rule, and the Subtraction Rule. Quotient is for division problems right? You can use the Quotient Rule whenever you have a fraction. If you take that approach, you will need the Quotient Rule and the Subtraction Rule. This is just subtraction?

Two points.

First, there is frequently more than one way to skin a cat (if you happen to enjoy that sort of thing).

Second, in many cases, finding the derivative involves combining a number of rules. You seem to be assuming that each derivative takes one rule to solve. Not so.

 

[Silvanoshei]

Ok, a little confusing, so I'll work here and hopefully that'll make this easier to understand. (Still foggy on why 5/(2s)=(5/2)s-² ... wouldn't it be (5/2)s^(-1)?)

So we have r'(s)= d/ds (3s)-² - d/ds(5/2)s-²
= 3 d/ds s-² - (5/2) d/ds s-²
= 3(-2)s-³ - (5/2)*(-2)s-³
= -6s-³ + (9/2)s-³ ?

 

[Deleted member 4993]

Finding the first and 2nd derivatives here.

How would I start finding the 1st derivative of r = 1 / 3s² - 5 / 2s ?

I tried putting them in neg. exponent form, like r = 3s-² ... but that doesn't work for 5/2s? Do I need to use the Product Rule? Quotient is for division problems right? This is just subtraction?

Is your original problem:

\(\displaystyle r \ = \ \frac{s^2}{3} \ - \ \frac{5s}{2} \)

or

\(\displaystyle r \ = \ \frac{1}{3s^2} \ - \ \frac{5}{2s} \)

The way you wrote it - it implied the first equation.

If you meant to write the second equation, you should have written:

r = 1 / (3s²) - 5 / (2s)

Please confirm the correct problem.

 

[Silvanoshei]

Sorry it is:

r = 1 / (3s²) - 5 / (2s)

 

[Deleted member 4993]

Sorry it is:

r = 1 / (3s²) - 5 / (2s)

r = 1/3 * s^-2 - 5/2 * s^-1

Now apply

\(\displaystyle \frac{d(x^n)}{dx} \ = \ n * x^{n-1}\)

and apply again for second derivative

 

[JeffM]

Ok, a little confusing, so I'll work here and hopefully that'll make this easier to understand. (Still foggy on why 5/(2s)=(5/2)s-² ... wouldn't it be (5/2)s^(-1)?) Yes it would. I made a mistake. See below

So we have r'(s)= d/ds (3s)-² But - d/ds(5/2)s-²
= 3 d/ds s-² - (5/2) d/ds s-²
= 3(-2)s-³ - (5/2)*(-2)s-³
= -6s-³ + (9/2)s-³ ?

Let's get you started correctly. Sorry for the blunder.

\(\displaystyle r = \dfrac{1}{3s^2} - \dfrac{5}{2s} = \frac{1}{3}s^{-2} - \frac{5}{2}s^{-1}.\) Now apply power and subtraction rules

\(\displaystyle \dfrac{dr}{ds} = \left(\frac{1}{3} * (- 2) * s^{-3}\right) - \left(\frac{5}{2} * (-1) * s^{-2}\right) = \dfrac{5}{2s^2} - \dfrac{2}{3s^3}\)