# Differentiate the functions.

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• Oct 8th 2012, 06:02 PM
calculus123
Differentiate the functions.
I was absent for the first time in my math class due to an emergency and now I'm falling behind everyone else in the class. I really need some help differentiating these functions. Please help me!

http://i49.tinypic.com/2clloz.jpg
http://i48.tinypic.com/2199c1g.jpg
• Oct 8th 2012, 07:02 PM
MarkFL
Re: Differentiate the functions.
Let's take these one at a time. First, it would be helpful to know what your thoughts are on how to differentiate the function in part a). Do you know the power rule? Do you know how to differentiate trigonometric functions?
• Oct 8th 2012, 07:21 PM
calculus123
Re: Differentiate the functions.
The power rule is http://mathworld.wolfram.com/images/...dEquation1.gif
How would I use this to solve a)?

Thanks Man!
• Oct 8th 2012, 07:24 PM
wtrmlncrawl191
Re: Differentiate the functions.
These problems are all just straight definitions. Use the power rule to do 2x^4 and -x^-4. Sec x and cos x are identities. There should be a table in your book. :)
• Oct 8th 2012, 07:24 PM
MarkFL
Re: Differentiate the functions.
You can apply the power rule to the first and last terms. Do you know what the derivatives of the secant and cosine functions are?
• Oct 8th 2012, 07:28 PM
calculus123
Re: Differentiate the functions.
The derivatives of the secant and cosine are SecXTanX and -Sinx, correct? I'm still very confused about approaching this problem. This might seem easy to you guys because you are all math geniuses here but I have a hard time with math in general.
• Oct 8th 2012, 07:34 PM
MarkFL
Re: Differentiate the functions.
Hey, we were all students once, we all had to learn, and you are doing fine. You have correctly given the power rule, and the rules for the two trig. functions involved, so now you just need to put it all together, term by term. What is your result?
• Oct 8th 2012, 07:41 PM
wtrmlncrawl191
Re: Differentiate the functions.
You need to work the problems step by step.

We can take the derivative of each individual portion.

f'(x) = (2x^4)' + (5sec(x))' - (3cos(x))' - (x^-4)'
Power rule: (2x^4)' = 8x^3
Secant Function: (5sec(x))' = 5sec(x)tan(x)
Cosine Function: (3cos(x))' = -3sin(x)
Power rule: (x^-4)' = -4x^-5

Now, sub back into the formula.
f'(x) = 6x^3 + 5sec(x)tan(x) + 3sin(x) + 4x^-5

If you are having trouble with the individual portions, that is mostly a memorization task. Just study the properties of derivatives.

Don't feel dumb. Really, everyone had to learn this at some point. It wasn't easy at first.
• Oct 8th 2012, 07:49 PM
MarkFL
Re: Differentiate the functions.
Quote:

Originally Posted by wtrmlncrawl191
...
Power rule: (2x^4)' = 6x^3...

You want:

$\frac{d}{dx}(2x^4)=8x^3$
• Oct 8th 2012, 07:54 PM
wtrmlncrawl191
Re: Differentiate the functions.
Guess I better stop giving advice past 10 pm...Oops. :/
• Oct 9th 2012, 11:42 AM
alane1994
Re: Differentiate the functions.
Quote:

Originally Posted by wtrmlncrawl191
Guess I better stop giving advice past 10 pm...Oops. :/

I am like that too, once it reaches a certain time of night, I am done.
Also, if you want a reference page try this one calculus. Mathwords: Derivative Rules
It has all of the basic rules on it, it is very helpful.:)
• Oct 10th 2012, 05:34 PM
calculus123
Re: Differentiate the functions.
Are these two correct?
http://i47.tinypic.com/pcgso.jpg
• Oct 10th 2012, 06:43 PM
MarkFL
Re: Differentiate the functions.
All but the first term in part b) are correct.
• Oct 10th 2012, 06:47 PM
calculus123
Re: Differentiate the functions.
You mean the 1/3x^2 ? I don't understand why it's wrong.
• Oct 10th 2012, 06:51 PM
wtrmlncrawl191
Re: Differentiate the functions.
Remember that 1/x^3 is the same as x^(-3). Then apply the power rule. The coefficient -3 goes on top, and the x^4 goes on the bottom.
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