# Checking a derivative

1. Jun 13, 2004

Am I on the right track?

Use the Summation Rule to find f ' (x) and simplify where possible:

f ' (x) = $$\frac {\sqrt {x}} {3} - \frac {3} {\sqrt {x}} + \frac {2} {x^3}$$

= $$\frac {1}{3} (\frac {1}{2} x^{-\frac {1}{2}}) - 3(-\frac {1}{2} x^{-\frac{3}{2}}) + 2(-3{x^{-4}})$$

= $$\frac {1} {6\sqrt {x}} + \frac {3} {2x^{\frac{3}{2}}} - \frac{6} {x^{4}}$$

2. Jun 13, 2004

### arildno

Seems OK to me

3. Jun 13, 2004

Derivative

Using the summation rule, find f ’(x) and simplify where possible:

f (x) = $$\frac {1}{(2x)^{3}} + \frac {\sqrt {x}}{2\sqrt [3] {x}} + \frac {x} {\sqrt {2}}$$

= $$\frac {1}{8}(-3)x^{-4}+ (\frac {1}{2})(\frac {1}{6})x^{-\frac {7}{6}}+\frac{1}{\sqrt{2}}$$

= $$\frac {-3}{8x^{4}} + \frac{1}{\sqrt{2}}$$ +1/12x^7/6

(Sorry about the notation couldn't get the last part right)

4. Jun 13, 2004

### arildno

You should have
$$\frac{1}{12}x^{-\frac{5}{6}}$$
rather than
$$\frac{1}{12}x^{-\frac{7}{6}}$$

otherwise OK.

5. Jun 13, 2004

Thanks..can you check this one?

Using the summation rule, find f ’(x) and simplify where possible:

F (x) = $$\frac {x^{4}+2x^{2}-3x}{4\sqrt{x}}$$

F ‘ (x) = $$(\frac{1}{4})(\frac{7}{2})x^{\frac {5}{2}}+ (\frac {1}{2})(\frac {3}{2})x^{\frac {1}{2}}-(\frac {3}{4})(\frac {1}{2})x^{-\frac{1}{2}}$$

= $$\frac {7}{8}x^{\frac{5}{2}}+\frac{3\sqrt{x}}{4}-\frac{3}{8\sqrt{x}}$$

6. Jun 13, 2004

### arildno

It seems you are ready to move on to more challenging problems, ladyrae; you're mastering these concepts now

7. Jun 13, 2004

Thanks...the next one is the Product Rule

Use the product rule to find f ‘ (x) and simplify where possible

f (x) = $$(x+e^{x})(3-\sqrt{x})$$

f ‘ (x) = $$(x+e^{x})(-\frac{1}{2}x^{-\frac{1}{2}})+(3-x^{\frac{1}{2}})(1+e^{t})$$

= $$-\frac{3}{2}x^{\frac{1}{2}}-\frac{e^{x}}{2x^{\frac{1}{2}}}+3+3e^{t}-x^{\frac{1}{2}}e^{t}$$

Is this right?

8. Jun 13, 2004

### Math Is Hard

Staff Emeritus
Which book are you working out of? Does it not offer the solutions - or a solution manual that you can buy separately? If you are using Stewart Calculus, I can show you where to download some of the solutions manual chapters for free.(At least I think the link's still there).

9. Jun 13, 2004

### Math Is Hard

Staff Emeritus
p.s. Wow! You are getting quite expert at the Latex formatting!!!

10. Jun 13, 2004

Thanks

I'm doing a calculus assignment...

Can anyone check the last one for me?

Thanks

11. Jun 13, 2004

### Math Is Hard

Staff Emeritus
where did the e^t come from that I see in your derivative?

12. Jun 13, 2004

terrible typo...i was reviewing another problem like it wrote down the wrong equation...

f (x) = $$(x+e^{x})(3-\sqrt{x})$$

f ‘ (x) = $$(x+e^{x})(-\frac{1}{2}x^{-\frac{1}{2}})+(3-x^{\frac{1}{2}})(1+e^{x})$$

= $$-\frac{3}{2}x^{\frac{1}{2}}-\frac{e^{x}}{2x^{\frac{1}{2}}}+3+3e^{x}-x^{\frac{1}{2}}e^{x}$$

Is this right?[/QUOTE]

13. Jun 13, 2004

2 left

Use the quotient rule to find f ’(x) and simplify where possible

Y = $$\frac{x^{3} + x} {x^{4}-2}$$

Y ‘ = $$\frac {(x^{4}-2) (3x^{2}+1)-(x^{3}+x)(4x^{3})}{x^{4}-2}$$

=$$\frac {-x^{6}-3x^{4}-6x^{2}-2}{x^{4}-2}$$

Is this right? How about the last one?

Anyone...Thanks

14. Jun 13, 2004

### arildno

Post 12 is correct.

The second line in 13 has the correct numerator, but you have forgotten to square the denominator.

15. Jun 13, 2004

thanks

Thanks..it should be

$$\frac {-x^{6}-3x^{4}-6x^{2}-2}{(x^{4}-2)^{2}}$$

can this be simplified?

16. Jun 13, 2004

last one

Find the equation of the tangent line to the curve $$y = x^{3}-1$$
at the point (-1,-2)

y ' = $$\frac {d}{dx}(x^{3}-1) = 3x^{2}$$

$$m = 3x^{2} = 3$$

$$(y-y_1) = m(x-x_1)$$

$$-3x+y-1 = 0$$

or
$$y = 3x + 1$$

17. Jun 13, 2004

### TALewis

If you have a TI-89, it will find all of these derivatives for you. It's a very nice way to check your work.

18. Jun 13, 2004

last one

Find the equation on the tangent line to the curve y = $$x^{3}-1$$at the point (-1,-2)

y ' = $$\frac {d}{dx}(x^{3}-1)=3x^{2}$$

$$m= 3x^{2} = 3$$

$$(y-y_1) = m(x-x_1)$$

$$-3x + y - 1 = 0$$

or
$$y=3x+1$$

is this right?

19. Jun 13, 2004

sorry

sorry for the repost..i didn't see page 2

Anyone?

20. Jun 13, 2004

### Math Is Hard

Staff Emeritus
You might also like to check out this site:

http://www.calc101.com

You can check derivatives here - although sometimes the steps it shows you will be more complicated than what you need to do to solve.