# Find the differential

• Mar 11th 2013, 02:30 PM
dokrbb
Find the differential
I really can not figure out how to proceed, I tried to solve it but apparently the solutions were wrong,

can someone give me a hint, please...

Let https://webwork.mathstat.concordia.c...4fe5928731.png.
Find the differential https://webwork.mathstat.concordia.c...e5fc553981.png when https://webwork.mathstat.concordia.c...556151e691.png and dx = 0.3

thank you
• Mar 11th 2013, 03:18 PM
HallsofIvy
Re: Find the differential
First, what did you get for the derivative of y with respect to x?
• Mar 11th 2013, 04:18 PM
dokrbb
Re: Find the differential
Quote:

Originally Posted by HallsofIvy
First, what did you get for the derivative of y with respect to x?

I tried this way f'(x)= sec^2(4x+7)(4)= (1/cos^2(4x+7))(4)
• Mar 11th 2013, 04:29 PM
Prove It
Re: Find the differential
It always helps if we can actually see what the pictures you post are and if their links aren't considered unsafe by the computer...
• Mar 11th 2013, 04:58 PM
dokrbb
Re: Find the differential
what do you mean?
• Mar 12th 2013, 03:27 PM
dokrbb
Re: Find the differential
Quote:

Originally Posted by Prove It
It always helps if we can actually see what the pictures you post are and if their links aren't considered unsafe by the computer...

oh, I got it, sorry for that, I thought it would be copied in LaTeX, it's from my univ official webpage, don't worry,

I still wait some help on it
• Mar 12th 2013, 05:52 PM
dokrbb
Re: Find the differential
Is my question so dumb, that's why it's ignored few days in a row?

Quote:

Originally Posted by dokrbb
I tried this way f'(x)= sec^2(4x+7)(4)= (1/cos^2(4x+7))(4)

• Mar 13th 2013, 03:38 AM
Prove It
Re: Find the differential
• Mar 13th 2013, 05:52 AM
dokrbb
Re: Find the differential
Quote:

Originally Posted by Prove It

I didn't expect you can`t see them, sorry for that.

The question is:
Let y = tan(4x + 7), find the dy when x= 2 and dx=0.3

I proceded by obtaining the derivative, but I`m completely confused about what should I do further:

f'(x)= sec^2(4x+7)(4)= (1/cos^2(4x+7))(4)...
• Mar 13th 2013, 07:36 AM
HallsofIvy
Re: Find the differential
Whoever gave you this problem expects you to know the basic definition of "differential"- if y= f(x) then dy= f'(x)dx.
• Mar 13th 2013, 08:01 AM
dokrbb
Re: Find the differential
Quote:

Originally Posted by HallsofIvy
Whoever gave you this problem expects you to know the basic definition of "differential"- if y= f(x) then dy= f'(x)dx.

I know the basic deffinition of the differential, I just don't know how to apply it in the case of a trigonometric function

I tried to solve it, f'(2) = (1/cos^2(4(2)+7))(4) =>

= (1/cos^2(15))(4) = ... and I recognize I have no idea how to deal with the trig expression cos^2(15)), do I have to consider it as cos(225), which gives me -0,707106, and I just have to plug in the values... (sorry, I haven't done any math since 2002, and I revised a bit of basic trigonometry, but I'm still lost in the trig identities and calculations. I hope though, I would manage to improve that by the time of my final exam...)
• Mar 13th 2013, 08:09 AM
HallsofIvy
Re: Find the differential
That's very basic algebra. No, (cos(x))^2 is NOT cos(x^2). It means exactly what it says- to find (cos(15))^2, find cos(15) (15 radians) and square that.
• Mar 13th 2013, 08:24 AM
dokrbb
Re: Find the differential
OK, in this case, I would have

f'(2) = (1/cos^2(4(2)+7))(4) =>

= (1/cos^2(15))(4)
= (1/(.96592582)^2)(4)
= (1/(.933012)(4)
= 4.287190;

after that, f'(2) = 4.287190 *(.3) = 1.286157, that's it?
• Mar 13th 2013, 09:42 AM
dokrbb
Re: Find the differential
Quote:

Originally Posted by dokrbb
OK, in this case, I would have

f'(2) = (1/cos^2(4(2)+7))(4) =>

= (1/cos^2(15))(4)
= (1/(.96592582)^2)(4)
= (1/(.933012)(4)
= 4.287190;

after that, f'(2) = 4.287190 *(.3) = 1.286157, that's it?

it still gives me a wrong answer, where is my mistake?
• Mar 13th 2013, 10:31 AM
HallsofIvy
Re: Find the differential
You are taking the cosine of 15 degrees when it should be 15 radians. Put your calculator into radian mode.