# Math Help - Help - Basic functions

1. ## Help - Basic functions

Can someone give me some guidance please, just started new calculas assignment and need some help:

If f(theta) = 10Sin^2(0.4theta - 0.1)
Show that f'(theta) = 4Sin(0.8theta - 0.2)

and evaluate f''(pie/4) to 3 decimal places?

For first part do I first change sin to cos and then remove the brackets which would give me 4Cos^2 but not much help with the rest?
I should get further than this but just need some help getting started.

Any help much appreciated.

2. Hi

$f(\theta) = 10 \sin^2(0.4 \theta - 0.1)$

$f(\theta) = 10 (\sin(0.4 \theta - 0.1))^2 = 10 \:u(\theta)^2$ where $u(\theta) = \sin(0.4 \theta - 0.1)$

The derivative of $(u(\theta))^2$ is $2\:u(\theta)\;u'(\theta)$

Therefore you have to find the derivative of $u(\theta) = \sin(0.4 \theta - 0.1)$

The derivative of $f(g(\theta))$ is $f'(g(\theta))\:g'(\theta)$

Can you go on now ?

3. ## ??

Sorry but you have lost me now. I might have missed something, why use u ang g? I thought to get the solution i needed to differentiate into dy/dx= and so find f' =??

4. I am sorry if I lost you ...

You are right : you need to differentiate f with respect to $\theta$ in order to find $f'(\theta)$

I used u, f and g to help but it seems that I missed the objective !
BTW I should have used g and h since f is already taken
$u(\theta) = \sin(0.4 \theta-0.1) = g(h(\theta))$ where $g(\theta)=\sin \theta$ and $h(\theta)=0.4 \theta-0.1$
$u'(\theta) = g'(h(\theta)) \:h'(\theta)$

u, g and h are intermediate functions aimed at using the usual formulas, that's all

5. ## Thanks

Thanks,

Makes it a bit more clearer, I can see where the 0.8 and 0.2 come from but how does the 10sin^2 become 4sin, I thought this was by multiplying out the brackets but I'm not sure now?

6. The differentiation of $u(\theta) = \sin(0.4 \theta - 0.1)$ is $u'(\theta) = 0.4\:\cos(0.4 \theta - 0.1)$

Therefore the differentiation of $(\sin(0.4 \theta - 0.1))^2$ is
$2\:\sin(0.4 \theta - 0.1)\:\:0.4\:\cos(0.4 \theta - 0.1)=0.4\:\:2\:\sin(0.4 \theta - 0.1)\:\cos(0.4 \theta - 0.1)$

Using trig identity $\sin(2x)=2\:\sin x\:\cos x$

The differentiation of $(\sin(0.4 \theta - 0.1))^2$ is $0.4\:\sin(0.8 \theta - 0.2)$

Now you just need to multiply this result by 10 to get the derivative of f

I hope this makes sense