C3 Differentiation Urgent Help needed!!!

The numerator would have to be positive for dy/dx to be positive, the denominator can be either positive or negative as squaring either a +ve or -ve number is still +ve?

Correct!

So you would say that the numerator has to be >0, right?

Now how does that example relate to the question at hand?

I've just realised that it doesn't technically matter as the square root of the denominator will never be negative, but it can be in some questions and you can still apply the same logic regardless.

Well so I have 1/2 + cos(x/2)>0, so x/2=cos^-1(-1/2), x=4.1887...

That's the critical value, so how do you find the range of values (remember the domain given in the question!)

Well when I've done these types of questions I usually get 2 critical values and then sketch a graph to see where the function is increasing/decreasing, but I only have one here

Remember that 0<x<2pi.so 0<x/2<pi. So you're looking of values for cosx/2 > -1/2 between 'actual inputs' of 0 and pi. So if you sketch a graph of cos between 0 and pi and look for y = -1/2, what do you notice?

I'm got a graph of y=cos x on my calculator and the x values for y= -1/2 are:

x=2.094... and x=4.1887...

You've drawn cosx between 0 and 2pi, but we're not interested in anything past pi because if x/2 > pi then x > 2pi, but in the question it states that 0 < x < 2pi.

Which is what I was saying in my previous post.

So is this just another trig question? So because I have x/2 instead of x I have to divide the domain by 2, so the new domain is 0<x/2<pi, and so I only use x/2=2.094... then I have x=4.1887...???

Yes, but the question doesn't want the critical point, it wants a range of values of x for which dy/dx > 0. So what is the range of values?

That's the bit I'm confused about..

Remember that 0<x<2pi.so 0<x/2<pi. So you're looking of values for cosx/2 > -1/2 between 'actual inputs' of 0 and pi. So if you sketch a graph of cos between 0 and pi and look for y = -1/2, what do you notice?

I can't say much more without giving it away or going in circles, but I'll try.



My advice would be to sketch (or use your calculator if you wish) costheta between 0 and pi, look at theta = 4pi/3 which is when costheta = - 1/2 (and since x/2 = theta, x = 8pi/3) and look at all the theta values to the left of that point and their y values - whether they are all greater than -1/2 or less than -1/2, etc.

Hmm but what do I write down? This question is worth 3 marks.

Please can someone help me with this question:

The function is y=e^x sin 2x. I've found dy/dx to be e^x(2cos 2x + sin 2x). I am asked to show that the equation of the normal to the curve y=e^x sin 2x at the point where
x=pi is 2e^(pi)y + x=pi.

I've tried plugging in x=pi into dy/dx to get e^pi(2cos2pi + sin2pi) which is the gradient of the tangent so the gradient of the normal is -1/e^pi(2cos2pi + sin2pi). I have tried many times already to rearrange to get the above expression but it's just not working out unfortunately. Please could someone help? Thanks.

My guess is 1st mark : recognising that it's the numerator that you're concerned with and rearranging.

2nd mark - finding the critical value.

3rd mark - writing down the correct answer, but you haven't found that yet.

But 4pi/3 isn't between 0 and pi right?

cos 2pi = 1, sin 2pi = 0.

Thanks

Correct. But x is between 0 and 2pi, and x = 4pi/3. x/2 is between 0 and pi, and cos(2pi/3) = -1/2.

So if x/2 is less than 2pi/3 then x is less than 4pi/3.