How to solve derivative of (sin3x)/x using first principle

First we must apply the Quotient rule that states,

This means that our derivative becomes,

Now we need to calculate and

From here the new equation looks like,

And that is the final result.

Hope this helps.

r3t40

Step-by-step explanation:

If , then  

.

To find this all I did was replace old input, x, with new input, x+h.

Now we will need this for our definition of derivative which is:

Before we go there I want to expand using the sum identity for sine:

So we could write f(x+h) as:

.

There are some important trigonometric limits we might need before proceeding with the definition for derivative:

Now let's go to the definition:

I'm going to clear the mini-fractions by multiplying top and bottom by a common multiple of the denominators which is x(x+h).

I'm going to distribute:

Now I’m going to group xsin(3x)cos(3h) with –xsin(3x) because I see when I factor this I might be able to use the second trigonometric limit I mentioned.  That is xsin(3x)cos(3h)-xsin(3x) can be factored as xsin(3x)[cos(3h)-1].

Now the limit I mentioned:

If I let u=3h then we have:

If 3h goes to 0, then h goes to 0:

If I multiply both sides by 3 I get:

I’m going to apply this definition after I break my limit using the factored form I mentioned for those two terms:

So the first limit I’m going to write as a product of limits so I can apply the limit I have above:

The first limit in that product of limits goes to 0 using our limit from above.

The second limit goes to sin(3x)/(x+h) which goes to sin(3x)/x since h goes to 0.

Since both limits exist we are good to proceed with that product.

Let’s look at the second limit given the first limit is 0. This is what we are left with looking at:

I’m going to write this as a sum of limits:

I can cancel out a factor of x in the first limit.  

I can cancel out a factor of h in the second limit.

Now I can almost use sin(u)/u goes to 1 as u goes to 0 for that first limit after writing it as a product of limits.  

The second limit I can go ahead and replace h with 0 since it won’t be over 0.

So this is what we are going to have after writing the first limit as a product of limits and applying h=0 to the second limit:

Now the first limit in the product I’m going to multiply it by 3/3 so I can apply my limit as sin(u)/u->1 then u goes to 0:

So we can plug in 0 for that last limit; the result will exist because we do not have over 0 when replacing h with 0.

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when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: