Differentiating Trignometric Functions

**sweetG** · July 19th 2008, 08:12 AM

I am REALLY not very good at this because I was away when they were teacher it at school and so I really need to brush up on this but anyway here are a few questions and could someone please help out I'd really appreciate it

Differentiate:

a) x^2 sin (-x)

b) sin^2 x

c) cosx.tan 3x

d) x/tan x

e) x^3 - 2 sin (x/3)

I'm sorry there are many questions here but I really don't know how to do these that well and I was away when they taught it at school

**Matt Westwood** · July 19th 2008, 12:19 PM

Start with (d) which is the most straightforward.

The quotient rule states:

$<br /> d \left({\frac u v}\right) = \frac {v du - u dv}{v^2}<br />$

Here $u = x, v = \tan x$

So $\frac {du}{dx} = 1, \frac {dv}{dx} = \sec^2 x$

This gives us $\frac d {dx} \left({\frac x {\tan x}}\right) = \frac {\tan x - x \sec^2 x}{\tan^2 x}$ which can of course be simplified.

The others involve the chain rule, which you'd best look up in the book because you need your wits about you.

You must have been away from class a long time, because this subject takes up quite a few lessons.

The chain rule says:

Let $y = f(u), u = g(x)$ .

Then $\frac {dy}{dx} = \frac {dy}{du} \frac {du}{dx}$ .

Start with (b) because it's easiest.

Here $u = \sin x$ and $y = u^2$ .

So $\frac {dy}{du} = 2 u = 2 \sin x$ and $\frac {du}{dx} = \cos x$ .

So $\frac {dy}{dx} = \frac {dy}{du} \frac {du}{dx} = 2 \sin x \cos x$ .

The others are a combination of chain rule and product rule (except (e) which is the sum of two separate terms which is easy).

Oh yeah, the product rule:

$d \left({uv}\right) = u dv + v du$ .

**tutor** · July 19th 2008, 07:17 PM

Hi,

a.)

x^2sin(-x)

here u have to apply product rule formula

=x^2d/dx(sin(-x))+sin(-x)d/dx(x^2)

note sin(-x)=-sin(x)

=-x^2cosx+-sinx(2x)
=-x^2cosx-2xsinx is the answer

b.)

sin^2x

=2sinxd/dx(sinx)
=2sinxcosx
=sin2x

c.)

cosxtan3x
=cosxd/dxtan3x+tan3xd/dxcosx
=cosx(sec^2(3x))d/dx(3x)+tan3x(-sinx)
=3cosxsec^2(3x)-sinxtan3x

d.)

x/tanx

here appllying the quotient rule formula

e.)

x^3-2sin(x/3)
=3x^2-2/3cos(x/3)

**Chris L T521** · July 19th 2008, 09:53 PM

Originally Posted by sweetG

I am REALLY not very good at this because I was away when they were teacher it at school and so I really need to brush up on this but anyway here are a few questions and could someone please help out I'd really appreciate it

Differentiate:

a) x^2 sin (-x)

Note that we can rewrite $\sin(-x)$ as $-\sin(x)$ due to the fact that sine is an odd function. Then apply product rule to $-x^2\sin(x)$ . Now that you know these derivatives better, it should be a breeze!

b) sin^2 x

Apply the chain rule. The way that Matt Westwood computed the derivative is great

c) cosx.tan 3x

Apply the product rule. Note that the chain rule is required when we end up differentiating $\tan({\color{red}3x})$

d) x/tan x

The way that Matt Westwood computed the derivative is perfectly fine. However, you can rewrite the equation as $x\cot(x)$ . You can use product rule here. Note that his answer and the answer you will get will be the same, but in alternative forms.

e) x^3 - 2 sin (x/3)

This should be a breeze for you now, but keep in mind that you will need to apply the chain rule when you differentiate $2\sin\left(\frac{x}{3}\right)$ .

I'm sorry there are many questions here but I really don't know how to do these that well and I was away when they taught it at school

I hope this makes more sense to you now!

--Chris

**Matt Westwood** · July 19th 2008, 10:37 PM

There is another way to look at $\sin \left({-x}\right)$ .

Rather than having to "know" that $\sin x$ is an "odd function", you can use the chain rule again:

.

Then

.

Here $u = -x$ and $y = \sin u$ .

Thus $\frac {du}{dx} = -1$ and $\frac {dy}{du} = \cos x$ .

This then leads to $\frac {dy}{dx} = \left({-1}\right) \cos x$ and there's the result.

It's a useful time-saver to learn that $\frac {d}{dx} \sin \left({nx}\right) = n \cos \left({nx}\right)$ , where $n$ is any number. (Similar results are there for all functions, not just sine.) This of course can be proved directly using the chain rule.

But it's interesting looking at a graph of $\sin \left({nx}\right)$ . It looks the same as $\sin x$ but you've got $n$ cycles of the sine squashed into the space for just one cycle. What you then see is that the slope of the sine wave has to be $n$ times steeper because it has to go up and down that many more times in that distance.

The above is seriously non-rigorous, but it may help you get an intuitive feel for what's going on in this case. Hope it helps.

Best of luck with the chain rule and product rule and all that - it's difficult getting to grips with the "why" of it, so it may be easier at this stage just to "learn the rules" and apply them mechanically. However, it's good to find out why it works, because that helps big time when you move on to integral calculus.

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