Thread: Derivative help

Have to find derivatives of following problems:




For this one, I simplified to :

(-4x^5 * x^(1/2)) + ((-7)/(x^3 * x^1/2))

-4^(11/2) + -7x^(-7/2)

Ended with:

-22x^(9/2) + (-49/2*x^(-9/2))

Is there something I'm doing wrong?




For this one I tried making it :

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

-15^-3/2 * t^-4

ended with 1/(15^(3/2) * t^4)

Is that correct?



I ended with 4/(x^(1/2))

Correct?



For this one, how would I do about solving this with the x on the bottom also?



Lastly:


How do you find the derivative at that point?

I got -6x-3 for the line, but what do i do after that?

For the first one, I think the second term should be positive (-7 * -7/2 = 49/2).

in future, don't bump

Originally Posted by FrustratedCollegeKid

Have to find derivatives of following problems:




For this one, I simplified to :

(-4x^5 * x^(1/2)) + ((-7)/(x^3 * x^1/2))

-4^(11/2) + -7x^(-7/2)

Ended with:

-22x^(9/2) + (-49/2*x^(-9/2))

Is there something I'm doing wrong?

The first term is correct, but the second term is not.

Originally Posted by FrustratedCollegeKid

For this one I tried making it :

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

-15^-3/2 * t^-4

ended with 1/(15^(3/2) * t^4)

Is that correct?

It's hard to read your results, but I do not think it is correct.

Originally Posted by FrustratedCollegeKid

I ended with 4/(x^(1/2))

Correct?

No, this one is wrong. If you've learned the chain rule, it is helpful to just memorize the derivative of the square root as a separate rule, since it comes up so often. That is,



If you haven't learned the chain rule yet, then do this as the power rule:

Originally Posted by FrustratedCollegeKid

For this one, how would I do about solving this with the x on the bottom also?

You need to use the quotient rule. Refer to your textbook, or this site:
Quotient Rule for Derivatives - HMC Calculus Tutorial

Originally Posted by FrustratedCollegeKid

How do you find the derivative at that point?

I got -6x-3 for the line, but what do i do after that?

Your formula for the derivative is correct. Now, you are given a pair of values in form. If you plug 5 into f(x), you'll see that you get -79 "out", or more formally we have that f(5)=-79. This is a point on the curve given by that equation. Now they want to know what the value of the derivative is at this point, that is, the point where x=5. So, you have , and you want to know what is...