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Derivative of Sine Function and Use of Chain Rule

Date: 11/06/2003 at 04:05:55
From: Courtney
Subject: differentiating a sine function

How do I differentiate a sine function?  I have learned about the 
product and chain rules but in my assignment I have to differentiate
f(x) = 2sin(0.2(x - 5)).

The sine part is the most confusing, but I'm also not sure whether to
use the product or chain rule.

Date: 11/06/2003 at 12:09:00
From: Doctor Riz
Subject: Re: differentiating a sine function

Hi Courtney -

Thanks for asking your questions.  Let's talk about differentiating
the sine function first.

Keep in mind what a derivative tells you - the slope of a tangent line 
to the original function at any point on the function.  If you think 
that way, you can probably figure out what the derivative of sin(x) is.

Visualize the graph of y = sin(x).  As you know, it starts at 0 when x 
is 0 degrees, then rises to 1 at 90 degrees, falls back to 0 at 180 
degrees, bottoms out at -1 at 270 degrees and returns to 0 at 360 
degrees, thus completing one full period of the curve.

Now think about the slope of the tangent line to the curve at each of 
those points.  At 0 degrees, the slope of the tangent is 1 since the 
curve is heading upwards at a 45 degree angle.  At 90 degrees, the
sine has hit a maximum, so the tangent line would sit right on top of
that 'hump' and have a slope of 0.  At 180 degrees, the sine is again
moving at an angle of 45 degrees, but it's now going downwards so the
slope of the tangent is -1.  At 270 degrees, the sine is at its
minimum value, so again the tangent line would be horizontal and have
a slope of 0.

Of course, between those 'key' points the slopes of the tangents are 
changing smoothly.  In other words, as the sine curve moves from 0 
degrees to 90 degrees, the tangent slopes are descending from 1 down
to 0 as the curve grows 'flatter'.

Now think about the four points we just came up with for values of the 
slope of the tangent line to the sine curve:

  0 degrees - slope of 1
  90 degrees - slope of 0
  180 degrees - slope of -1
  270 degrees - slope of 0
  360 degrees is the same as 0 degrees so again the slope is 1.

Can you think of another function that contains those points?  Here's
a hint - it's another trig funtion, and it perfectly shows the slope
of the sine curve at any point.  Thus, that other trig function is the 
derivative of the sine function.  I'll let you try to figure out what 
it is based on the numbers above.

You also asked about the chain rule versus the product rule.  Look at 
the function you are trying to differentiate:  

  f(x) = 2sin[0.2(x - 5)].

What is there for multiplication in this function?  The 2 out front
can be treated as multiplying by a constant and can just be left out
front.  The 0.2 inside the sine function is again just a constant (ie,
it's not 0.2x), so it doesn't need to be treated as its own function.
So, there really isn't any place in this where two functions of x are
being multiplied.  That means we won't need the product rule.

But there IS a place where there is a function INSIDE another
function, which is called a composition of functions.  In other words,
within the main function of sin(x) there is the function 0.2(x - 5). 
It might be easiest to just distribute that out, which gives 0.2x - 1.
So now we have:

  f(x) = 2sin(0.2x - 1)

Let's leave the 2 out front and take the derivative of the OUTSIDE 
function, which is the sin(x) part.  As you might have figured out
from the earlier discussion, the derivative of sin(x) is cos(x).  So
that gives us:

  f'(x) = 2 * cos(0.2x - 1)

Now the chain rule says to take the derivative of the INSIDE function, 
which is the (0.2x - 1) part.  The derivative of that is 0.2, so we 
multiply what we had already times 0.2.  All of that gives us:

  f'(x) = 2 * cos(0.2x - 1)*(0.2)

Multiplying the 2 and the (0.2) our final answer is 

  f'(x) = 0.4 * cos(0.2x - 1).  

Of course, if you want to make it look more like the original
function, you could change the (0.2x - 1) back into (0.2)(x - 5) and 
the derivative would become:

  f'(x) = 0.4cos[0.2(x - 5)]

Here's another example.  Suppose you want to differentiate 

  y = sin(x^2 + 3x)  

Start by taking the derivative of the sin to get cos(x^2 + 3x).  Now 
the chain rule kicks in to take the derivative of x^2 + 3x, which is
2x + 3.  So the final answer is 

  y' = cos(x^2 + 3x) * (2x + 3) or (2x + 3)cos(x^2 + 3x).

Here's a good question for you.  Can you figure out what the
derivative of cos(x) is by thinking in terms of slopes at the quarter
points like we did with sin(x)?  If you understood that part, you
should be able to figure out the derivative of cos(x).

One final comment - if you come upon something you don't know how to 
take the derivative of, most calculus texts have a table in the back
or sometimes inside the cover which shows many common derivatives, so
you might be able to look it up in a table.

Hope that helps.  Feel free to write back if you are still confused.

- Doctor Riz, The Math Forum 
Associated Topics:
High School Calculus

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