# Thread: Derivative help

1. ## 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?

2. ## Re: Derivative help

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

3. ## Re: Derivative help

$\displaystyle \frac{\sqrt{5}}{t^3} = \sqrt{5} \cdot t^{-3}$

$\displaystyle \frac{d}{dt} \left[\sqrt{5} \cdot t^{-3}\right] =$

$\displaystyle \sqrt{5} \frac{d}{dt} \left[t^{-3}\right] =$

$\displaystyle \sqrt{5} \left(-3t^{-4})$

$\displaystyle -\frac{3\sqrt{5}}{t^4}$

in future, don't bump

4. ## Re: Derivative help

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.

$\displaystyle \frac d {dx} (-7x^{-\frac 7 2}) = (-7) \frac d {dx} x^{-\frac 7 2}=-7 [(-\frac 7 2)x^{-\frac 7 2 - 1}] = \frac {49} 2 x^{-\frac 9 2}$

5. ## Re: Derivative help

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.
$\displaystyle \frac d {dx} \sqrt{5} t^{-3} = \sqrt{5} \frac d {dx} t^{-3} = \sqrt{5} [(-3)t^{-3 -1}] = -3\sqrt 5 t^{-4}$

6. ## Re: Derivative help

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,

$\displaystyle \frac d {dx} \sqrt{f(x)} = \frac {f'(x)} {2\sqrt{f(x)}}$

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

$\displaystyle \frac d {dx} \sqrt{8x} = \sqrt 8 \frac d {dx}x^{\frac 1 2} = \sqrt 8 [(\tfrac 1 2)x^{\frac 1 2 - 1}] = 2\sqrt 2 [\tfrac 1 2 x^{-\frac 1 2}] = \sqrt{\frac 2 x}$

7. ## Re: Derivative help

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

8. ## Re: Derivative help

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 $\displaystyle (x,y)$ 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 $\displaystyle f'(x)=-6x-3$, and you want to know what $\displaystyle f'(5)$ is...