Finding the slope of tangent line x = 30 degrees for f(x) = sin(x)

[PM123]

The problem is:

Find the slope of the tangent line at x = 30 degrees for f(x) = sin(x).

Using the derivatives formula, I got:
lim h->0 = [sin(x+h) - sin(x)]/h

If I plug in 30 for x I get [sin(30+0) - sin(30)]/0 which is basically 0/0. Usually that means you can do more within the problem, how do I continue?

Thanks!

 

[Dr.Peterson]

The problem is:

Find the slope of the tangent line at x = 30 degrees for f(x) = sin(x).

Using the derivatives formula, I got:
lim h->0 = [sin(x+h) - sin(x)]/h

If I plug in 30 for x I get [sin(30+0) - sin(30)]/0 which is basically 0/0. Usually that means you can do more within the problem, how do I continue?

Thanks!

There are a couple issues to deal with.

First, the derivative is the limit of the difference quotient; and you are right that the indeterminate form means that you have to more work to find the limit. I would not expect you to be asked to derive the formula for the derivative of the sine in the middle of a problem like this; are you sure you haven't been taught the formula? The derivation depends on first knowing that the limit of sin(x)/x is 1; have you at least seen that?

Second, we don't usually talk about the derivative of trig functions expressed in terms of degrees; differentiation works nicely only in terms of radians. (That is a large part of the reason we use radians.) If you really want the slope of the tangent to the function y = sin(x°), you need to do some extra work. This can involve treating it as the function y = sin(pi/180 x).

So it will be important for you to tell us the context of the question. If this is an exercise you were given, please state it completely, word for word, with anything else that was said leading up to it. And tell us what you have been taught about derivatives of trig functions in particular.

 

[j-astron]

The problem is:

Find the slope of the tangent line at x = 30 degrees for f(x) = sin(x).

Using the derivatives formula, I got:
lim h->0 = [sin(x+h) - sin(x)]/h

If I plug in 30 for x I get [sin(30+0) - sin(30)]/0 which is basically 0/0. Usually that means you can do more within the problem, how do I continue?

Thanks!

Hi PM123

A couple of points:

1) The argument that goes to a trigonometric function is always a dimensionless number (a number with no units). Think back to the original definitions of sine/cosine in terms of the unit circle to see why this is the only thing that makes sense. Therefore, the only type of angle you can supply as an argument to a trigonometric function is the angle in radians. Although a radian is a unit, it's a dimensionless one, because it is defined as the ratio of two lengths. Therefore angles in radians are just numbers. Writing \(\displaystyle \sin\left(30^\circ\right)\) is actually just a shorthand for writing \(\displaystyle \sin\left(30^\circ \cdot \frac{\pi}{180^\circ}\right)\) where the thing left in parentheses is a pure number, as it must be. Long story short: convert your angles to radians before passing them as arguments to trigonometric functions. Your calculator will do it for you if you're in degree mode, but if you're doing algebra you will get into trouble, is what I'm saying.

2) Do you really have to use the formal definition of a derivative in terms of limits, in order to solve this problem? Have you learned other techniques for differentiating functions yet? What function is the derivative of the sine function? If you don't know the answer to that yet, or the question doesn't make sense, then the only way I can think of for you to solve this problem is to plug in successively smaller and smaller values of h and see what your answer is tending towards. Remember that \(\displaystyle \lim_{h\rightarrow 0}f(h) \) does not mean the same thing as just \(\displaystyle f(h=0)\). The limit can exist even if the function is not defined there...it refers to what your values are tending towards as h tends towards 0.

EDIT: I think I said all the same things as DrPeterson, only slower and more verbosely, LOL

 

[sinx]

d/dx sinx

The problem is:

Find the slope of the tangent line at x = 30 degrees for f(x) = sin(x).

Using the derivatives formula, I got:
lim h->0 = [sin(x+h) - sin(x)]/h

If I plug in 30 for x I get [sin(30+0) - sin(30)]/0 which is basically 0/0. Usually that means you can do more within the problem, how do I continue?

Thanks!

d/dx sinx= cosx; but you can prove it thru the derivative formula.
expand sin(x+h) using the trig identity sin(a+b)=sina*cosb+cosa*sinb
then collect terms of the derivative formula into terms of sinx and cosx; then you have two fractions.

i.e. d/dx sinx=lim_h->0 [sin(x+h) - sin(x)]/h
=lim_h->0 sinx *[cosh-1]/h + cos x*[sinh]/h

[sinh]/h will need to be determined thru 'plug and see', using a table. You will see the fraction approaches 1 as h approaches 0.
[cosh-1]/h will need another identity, sinx/2=((1-cosx)/2)^1/2; or 1-cosh=2sin²h/2;

the result is d/dx sinx= cosx

then the slope of tangent line = cos30