Re: Find the differential

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

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)

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...

Re: Find the differential

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

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)

Re: Find the differential

It might be because your question CAN NOT BE READ.

Re: Find the differential

Quote:

Originally Posted by

**Prove It** It might be because your question CAN NOT BE READ.

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)...

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.

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...)

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**.

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?

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?

Re: Find the differential

You are taking the cosine of 15 **degrees** when it should be 15 **radians**. Put your calculator into radian mode.