Checking a derivative

[ladyrae]

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}}

[arildno]

Seems OK to me

[ladyrae]

How about this one? Is this right?

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)

[arildno]

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

otherwise OK.

[ladyrae]

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}}

[arildno]

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

[ladyrae]

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?

[Math Is Hard]

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).

[Math Is Hard]

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

[ladyrae]

I'm doing a calculus assignment...

Can anyone check the last one for me?

Thanks

[Math Is Hard]

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

[ladyrae]

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]

[ladyrae]

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

[arildno]

Post 12 is correct.

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

[ladyrae]

Thanks..it should be

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

can this be simplified?

[ladyrae]

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

[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.

[ladyrae]

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?

[ladyrae]

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

Anyone?

[Math Is Hard]

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.

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.

[Math Is Hard]

and y=3x+1 looks correct to me.