Trig Functions -- 2U Fitzpatrick 17(f) (1 Viewer)

[Pyramid]

So I got everything in the exercise right, except for the last 2 questions, as below:

Find the derivatives of:

Q44) tan(pi - x)

Provided Answer: -[sec(x)]^2

Q45) tanx`

Provided Answer: (pi/180)[sec(x')]^

I'd appreciate if someone could explain you get to these solutions. Thanks

Note:
- by "pi" I mean the 3.14 number...
- by " ' " on tanx' and [sec(x')]^2 I mean "degrees".

 

[frenzal_dude]

So I got everything in the exercise right, except for the last 2 questions, as below:

Find the derivatives of:

Q44) tan(pi - x)

Answer: -sec^2X

Q45) tanx`

Answer: (pi/180)sec^2x`

I'd appreciate if someone could explain you get to these solutions. Thanks

Note:
- by "pi" I mean the 3.14 number...
- by " ' " on tanx' and sec^2x' I mean "degrees".

hey,
if you write trig functions with a square in them pls write it like this: -[sec(x)]^2 and (pi/180)[sec(x)]^2 coz it looks clearer and easier to read and we know if the x is part of the exponent or not.

So for the first one:
tan(pi-x) you take the derivative of the angle argument which is -1, then you times that by the generic derivative of tan which is [sec(x)]^2 so you should get -[sec(pi-x)]^2

For the 2nd one:
tanx just goes to [sec(x)]^2

 

[Pyramid]

So for the first one:
tan(pi-x) you take the derivative of the angle argument which is -1, then you times that by the generic derivative of tan which is [sec(x)]^2 so you should get -[sec(pi-x)]^2

For the 2nd one:
tanx just goes to [sec(x)]^2

1) I got your answer, and by your same reasoning as well. So maybe the Fitzpatrick answer is wrong?

2) I'm not sure about your answer for this. Fitzpatrick's answer involves some manipulation as a result of x being in degrees. Besides, it's the last question of the exercise, it can't be as easy as your answer has made it out to be.

Thanks for your help, too bad about dodgy Fitzpatrick

 

[Pyramid]

[ is a 2nd quadrant angle, and tan is -ve in the 2nd quadrant]
Differentiate this, and you get the required answer

2. To convert an angle from degrees to radians, you multiply by
So
When you differentiate this, you get
The only reason for converting to radians was so that we could differentiate (the rules you've learned only apply when the angles are in radians). Now that we have differentiated, we might as well convert back - and we get the required answer.

Thankyou on both counts!