Question

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

Answer 1

$\dfrac{d}{dx}(\dfrac{\sin(3x)}{x})$

First we must apply the Quotient rule that states,

$(\dfrac{f}{g})'=\dfrac{f'g-g'f}{g^2}$

This means that our derivative becomes,

$\dfrac{\dfrac{d}{dx}(\sin(3x))x-\dfrac{d}{dx}(x)\sin(3x)}{x^2}$

Now we need to calculate $\dfrac{d}{dx}(\sin(3x))$ and $\dfrac{d}{dx}(x)$

$\dfrac{d}{dx}(\sin(3x))=\cos(3x)\cdot3$

$\dfrac{d}{dx}(x)=1$

From here the new equation looks like,

$\dfrac{3x\cos(3x)-\sin(3x)}{x^2}$

And that is the final result.

Hope this helps.

r3t40

Answer 2

Answer:

$\frac{3\cos(3x)}{x}-\frac{\sin(3x)}{x^2}$

Step-by-step explanation:

If $f(x)=\frac{\sin(3x)}{x}$, then  

$f(x+h)=\frac{\sin(3(x+h)}{x+h}=\frac{\sin(3x+3h)}{x+h}$.

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:

$f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Before we go there I want to expand $sin(3x+3h)$ using the sum identity for sine:

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

$\sin(3x+3h)=\sin(3x)\cos(3h)+\cos(3x)\sin(3h)$

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

$f(x+h)=\frac{\sin(3x)\cos(3h)+\cos(3x)\sin(3h)}{x+h}$.

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

$\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1$

$\lim_{u \rightarrow 0}\frac{\cos(u)-1}{u}=0$

Now let's go to the definition:

$f'(x)=\lim_{h \rightarrow 0}\frac{\frac{\sin(3x)\cos(3h)+\cos(3x)\sin(3h)}{x+h}-\frac{\sin(3x)}{x}}{h}$

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).

$f'(x)=\lim_{h \rightarrow 0}\frac{x(\sin(3x)\cos(3h)+\cos(3x)\sin(3h))-(x+h)\sin(3x)}{x(x+h)h}$

I'm going to distribute:

$f'(x)=\lim_{h \rightarrow 0}\frac{x\sin(3x)\cos(3h)+x\cos(3x)\sin(3h)-x\sin(3x)-h\sin(3x)}{x(x+h)h}$

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:

$\lim_{u \rightarrow 0}\frac{\cos(u)-1}{u}=0$

If I let u=3h then we have:

$\lim_{3h \rightarrow 0}\frac{\cos(3h)-1}{3h}=0$

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

$\lim_{h \rightarrow 0}\frac{\cos(3h)-1}{3h}=0$

If I multiply both sides by 3 I get:

$\lim_{h \rightarrow 0}\frac{\cos(3h)-1}{h}=0$

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

$f'(x)=\lim_{h \rightarrow 0}\frac{x\sin(3x)\cos(3h)-x\sin(3x)+x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}$

$f'(x)=\lim_{h \rightarrow 0}\frac{x\sin(3x)(\cos(3h)-1)+x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}$

$f'(x)=\lim_{h \rightarrow 0}\frac{x\sin(3x)(\cos(3h)-1)}{x(x+h)h}+\lim_{h \rightarrow 0}\frac{x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}$

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

$f’(x)=\lim_{h \rightarrow 0}\frac{\cos(3h)-1}{h} \cdot \lim_{h \rightarrow 0}\frac{x\sin(3x)}{x(x+h)}+\lim_{h \rightarrow 0}\frac{x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}$

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:

$f’(x)=\lim_{h \rightarrow 0}\frac{x\cos(3x)\sin(3h)-h\sin(3x)}{x(x+h)h}$

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

$\lim_{h \rightarrow 0}\frac{x\cos(3x)\sin(3h)}{x(x+h)h}+\lim_{h \rightarrow 0}\frac{-h\sin(3x)}{x(x+h)h}$

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

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

$\lim_{h \rightarrow 0}\frac{\cos(3x)\sin(3h)}{(x+h)h}+\lim_{h \rightarrow 0}\frac{-\sin(3x)}{x(x+h)}$

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:

$\lim_{h \rightarrow 0}\frac{\sin(3h)}{h} \cdot \lim_{h \rightarrow 0}\frac{\cos(3x)}{(x+h)}+\frac{-\sin(3x)}{x(x+0)}$

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:

$\lim_{h \rightarrow 0}3\frac{\sin(3h)}{3h} \cdot \lim_{h \rightarrow 0}\frac{\cos(3x)}{(x+h)}+\frac{-\sin(3x)}{x(x)}$

$3(1) \cdot \lim_{h \rightarrow 0}\frac{\cos(3x)}{(x+h)}+\frac{-\sin(3x)}{x(x)}$

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.

$3(1)\frac{\cos(3x)}{x}+\frac{-\sin(3x)}{x^2}$

$\frac{3\cos(3x)}{x}-\frac{\sin(3x)}{x^2}$