Mathematics Question Thread

[RuiAce]

(Image removed from quote.)

[Shadowxo]

PLEASE HELP

I already answered this question yesterday, next time please check if your question has been answered before reposting
If an answer doesn't explain it, say what you don't understand and we'll help!

[Mathew587]

Hey all
Everything is in the image
Thanks peeps :D

[anotherworld2b]

I'm not quite sure how to do these questions

[RuiAce]

Hey all
Everything is in the image
Thanks peeps

I'm not too sure what you did there. There's only one G.P. going on, not two.

[Mathew587]

I'm not too sure what you did there. There's only one G.P. going on, not two.

oh ok ty rui

[Shadowxo]

I'm not quite sure how to do these questions

18.That question's a bit of a tough one but I think this is how they want you to solve it.

[anotherworld2b]

thank you for your help

18.That question's a bit of a tough one but I think this is how they want you to solve it.

[Arisa_90]

Hi there
Could someone help me understand the rules for trig and exponential differentiation? I don't know how to use the rules properly so I keep getting questions wrong.
Could someone also explain to me how to draw gradient function graphs as well as drawing the graph given the gradient function?
I would like to make step by step notes on these things

[jamonwindeyer]

Hi there
Could someone help me understand the rules for trig and exponential differentiation? I don't know how to use the rules properly so I keep getting questions wrong.
Could someone also explain to me how to draw gradient function graphs as well as drawing the graph given the gradient function?
I would like to make step by step notes on these things

Hey Arisa! Welcome to the forums!

So the rules for trig and exponential differentiation are on your reference sheet, but they all just involve multiplying by the derivative of the inside. So for example, consider \(y=\sin{x^2}\).

The 'inside' bit of the function is \(x^2\), the derivative of which is \(2x\). So we'll be multiplying by that. Now normally, the derivative of \(\sin\) is \(\cos\), so put it all together, and you get \(y'=2x\cos{x^2}\). Notice that the inside bit doesn't change!

This reflects the rule on your reference sheet, \(\frac{d}{dx}\sin{f(x)}=f'(x)\cos{f(x)}\) - Swap to cos and multiply by the inside derivative!

To improve your working there, why don't you upload some questions you've been struggling with and your attempts at solving them - We can try to show you the way to tackle them

As for sketching gradient functions from regular graphs, again, it is just practice. But here are a few rules to help (possibly pop these in your notes):

- Maximum turning points turn into x-intercepts, going from positive to negative
- Minimum turning points turn into x-intercepts, going from negative to positive
- Inflexions turn into maxima/minima
- Intercepts are useless

I always approach these questions by marking the sign of the gradient at all points. That is, if it slopes up, mark with a '+'. If down, mark with a '-'. This will show you at a glance where the graph of your gradient function should be above the x-axis and where it should be below.

For going backwards, the rules are reversed a little:

- X-Intercepts turn into maxima/minima
- Turning points turn into inflexions
- Inflexions are useless

Again, the best way to improve here won't be to makes notes (though that is definitely a good thing to do). You need to practice!

Once again, welcome! Be sure to upload any specific questions you have so we can help you out!

[Arisa_90]

How do you know when its going from positive to negative and vice versa for the gradient function to the original function?

I found some examples for trig differentiation that I am having trouble with
1. y= tanx^2
2. y= tan^3(2x)
3. y= sin^2(1+squareroot t)
4. y= square root cos(e^2t)
5. y= e^sin3t
6. y= (1+2x)^3 tan(1-squareroot x)
7. y=sint/cos2t

Hey Arisa! Welcome to the forums!

So the rules for trig and exponential differentiation are on your reference sheet, but they all just involve multiplying by the derivative of the inside. So for example, consider \(y=\sin{x^2}\).

The 'inside' bit of the function is \(x^2\), the derivative of which is \(2x\). So we'll be multiplying by that. Now normally, the derivative of \(\sin\) is \(\cos\), so put it all together, and you get \(y'=2x\cos{x^2}\). Notice that the inside bit doesn't change!

This reflects the rule on your reference sheet, \(\frac{d}{dx}\sin{f(x)}=f'(x)\cos{f(x)}\) - Swap to cos and multiply by the inside derivative!

To improve your working there, why don't you upload some questions you've been struggling with and your attempts at solving them - We can try to show you the way to tackle them

As for sketching gradient functions from regular graphs, again, it is just practice. But here are a few rules to help (possibly pop these in your notes):

- Maximum turning points turn into x-intercepts, going from positive to negative
- Minimum turning points turn into x-intercepts, going from negative to positive
- Inflexions turn into maxima/minima
- Intercepts are useless

I always approach these questions by marking the sign of the gradient at all points. That is, if it slopes up, mark with a '+'. If down, mark with a '-'. This will show you at a glance where the graph of your gradient function should be above the x-axis and where it should be below.

For going backwards, the rules are reversed a little:

- X-Intercepts turn into maxima/minima
- Turning points turn into inflexions
- Inflexions are useless

Again, the best way to improve here won't be to makes notes (though that is definitely a good thing to do). You need to practice!

Once again, welcome! Be sure to upload any specific questions you have so we can help you out!

[Shadowxo]

How do you know when its going from positive to negative and vice versa for the gradient function to the original function?

I found some examples for trig differentiation that I am having trouble with
1. y= tanx^2
2. y= tan^3(2x)
3. y= sin^2(1+squareroot t)
4. y= square root cos(e^2t)
5. y= e^sin3t
6. y= (1+2x)^3 tan(1-squareroot x)
7. y=sint/cos2t

Hi Arisa
When going from the gradient function to the original function, you don't know if the values on the original function are positive or negative - as when integrating you add +c. The gradient function shows you whether the function is going up or down - when the gradient is negative (below 0 on the gradient function), the original graph is going downwards, and when the gradient is positive (above 0 on the gradient function) it's going up. So the gradient function can help you determine the shape of the original graph, but not how far up or down it's translated (unless you're given a point on the original function).

So for your examples, you find the derivative of the outside and multiply by the derivative of the inside. If we represent these functions as g(f(x)), the derivative is g'(f(x)) * f'(x)
1. The derivative of tan is sec², and the derivative of x² is 2x. So the derivative of tan(x²) is sec²(x²) * 2x. If you want to graph this gradient function, you can see it's always negative when x is less than zero (sec² is always positive) and positive when x>0, but the values fluctuate depending on the x value as cos²(x) is between 0-1
2. Same thing - derivative of tan³ is 3tan²*sec², and the derivative of 2x is 2. So the derivative is
3tan²(2x)*sec²(2x)*2

Try some of the other questions and let me know if there's anything more you want to know / want clarification on

[Hplovers]

Hi! I'd love your help, I keep getting confused with what do do when theres are area in the negative field of the graph...

19. Find the volume of the solid formed when the line x+3y-1=0 is rotated about the x-axis from x=0 to x=8.

I keep getting the answer but in negative form and not sure what to do about this as I could simply add absolute value brackets but I know thats not what your supposed to do.
Thanks!!

[RuiAce]

Hi! I'd love your help, I keep getting confused with what do do when theres are area in the negative field of the graph...

19. Find the volume of the solid formed when the line x+3y-1=0 is rotated about the x-axis from x=0 to x=8.

I keep getting the answer but in negative form and not sure what to do about this as I could simply add absolute value brackets but I know thats not what your supposed to do.
Thanks!!

A volume should never end up negative. If this happens, you should post up your working so we can look out for mistakes.

[cxmplete]

hi!
how would i do question 14, im so confused!