# Thread: Derivatives and Tangent Lines

1. ## Derivatives and Tangent Lines

Hey. I have a question here I was hoping to get a hand with.

Determine the slope of the tangent at x = 3 for the function :

f(x) = (4x^3) / sinx

Note -- x is measured in radians

I've gotten the derivative without a problem, it looks like this:

(12x^2)(1/sinx) - (4x^3) (1/(sinx^2))(cosx)

I think where I'm getting hung up is the radians issue. I know the pi rad = 180 deg... but could you help me figure this out. My book is giving (sinx) as equal to 0.1411. Thanks.

2. Hello jools. Looks like you have the derivative figured out just fine. So now simply plug the value 3 in for x. If your calculator requires the use of degrees for the argument of sin and cosine, simply convert 3 radians into degees by multiplying by 180/ $\pi$. You'll find that 3 radians is equivalent to about 171.9 degrees.

3. Most calculators have can be set to either "radians" or "degrees". Set your calculator to radians and you won't need to worry about converting. For any math course from Pre-Calculus up you will want your calculator in radian mode.

4. Got it! Thanks guys. So for these types of questions, will value of x always be measured in radians or only when it's noted in the question?

5. Unless stated otherwsie, assume that angles are in radians, especially if the problem involves trig functions. However, I must say it's rare to see an angle that is something other than a multiple of $\pi$.

6. Originally Posted by Jools
Got it! Thanks guys. So for these types of questions, will value of x always be measured in radians or only when it's noted in the question?
If you are usng calculus it is almost certain you'll want to work in degrees otherwise it get's a lot more complicated eg:

In radians if $f(x) = sin x$ then $f'(x) = cos x$

But in degrees if $f(x) = sin x$ then $f'(x) = \frac{\pi}{180}cos x$

7. Alright. So just to confirm I've got this. I did this question also:

Determine the slope of the tangent at x=0 for the function:

f(x) = (cosx) / (1-x)

I worked the derivative out to be:

(-sinx)(1-x)^-1 + (cosx)(-1)(1-x)^-2(-1)

or

(-sinx)/(1-x) + 1/(1-x)^2

= 0/(1-x) + 1/(1-1)^2

= 1

Have I got it? Thanks again.

8. You seem to have dropped the cos(x) term from the second half of the derivatve:

$
f'(x) = \frac {-sin(x)} {1-x} + \frac {cos(x)} {(1-x)^2}
$

So:

$
f'(0) = \frac {-sin(0)} {1-0} + \frac {cos(0)} {(1-0)^2} = 1.
$

9. Originally Posted by Jools
Alright. So just to confirm I've got this. I did this question also:

Determine the slope of the tangent at x=0 for the function:

f(x) = (cosx) / (1-x)

I worked the derivative out to be:

(-sinx)(1-x)^-1 + (cosx)(-1)(1-x)^-2(-1)

or

(-sinx)/(1-x) + 1/(1-x)^2

= 0/(1-x) + 1/(1-1)^2

= 1

Have I got it? Thanks again.
Perfect! You can also check these with a graphing calculator.

10. Originally Posted by mathemagister
Perfect! You can also check these with a graphing calculator.
Excellent. Thank you!

11. Originally Posted by Jools
Excellent. Thank you!
The blue line is the derivative. So look at the blue line at x=0 and you have your answer. You can check every question like this. [Like the saying: "Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime."]

Hope that helps you

12. Obviously it's nice to use the graphing calculator to check your work, but not as a substitute for doing the work in the first place.

13. Originally Posted by ebaines
Obviously it's nice to use the graphing calculator to check your work, but not as a substitute for doing the work in the first place.
Of course, since that would defeat the purpose. I always feel unsatisfied if, after doing a problem with a calculator, I realize I could have done it by hand.