TheAspiringDoc's Math Thread

[keltingmeith]

I'm aware that the derivative of trig functions like sin(x) is only cos(x) if x is in radians (right?), but does that also apply for inverse trig [ I.e. Does the derivative of cos^-1 (x) only end up being -1/sqrt(1-x^2) if x is in radians?]
Thank you

First point - correct.

Second point - not quite. It has something to do with the output instead. What is the relationship between sin(x) and arcsin(x)?

[MathsGuru]

I'm aware that the derivative of trig functions like sin(x) is only cos(x) if x is in radians (right?), but does that also apply for inverse trig [ I.e. Does the derivative of cos^-1 (x) only end up being -1/sqrt(1-x^2) if x is in radians?]
Thank you

As EulerFan101 pointed out, you're almost right - what you want to say is that the expression you gave for the derivative of y = cos^-1(x) depends on y, not x, being measured in radians.

Here's a nice way to think about the radian dependence geometrically. 

As the Methods textbooks will tell you (around early Yr12 material), the graph of an inverse function can be obtained by reflecting the original graph in the line y = x, so that x- and y-coords are swapped.

So, while changing the angle measurement unit dilates the graph horizontally for a trig function, it dilates it vertically for an inverse trig fn.  In either case, this dilation is going to stretch the rise or run of any triangle you use to measure or to approximate the gradient, by the same factor, so the numerical value of the derivative must also change accordingly.

[TheAspiringDoc]

The product rule is as follows:   d/dx (f/g) =  (f’ g − g’ f )/g^2

In the product rule as shown above, do 'f' and 'g' represent two (different?) functions of x?
So it'd then be better to write as:
?
thnx 

[lzxnl]

In the product rule as shown above, do 'f' and 'g' represent two (different?) functions of x?
So it'd then be better to write as:
?
thnx 

That's the quotient rule.

[TheAspiringDoc]

Hey,
So once you've learnt the product, quotient and chain rules, as well as all of the minor rules (e.g. d/dx [ln(x)] = 1/x or d/dx [tan^-1] = 1/(1+x^2)), what's the next thing to learn? Surely there's more to it, I just can't find 'what's next' I guess.. ?
 

[_fruitcake_]

Hey,
So once you've learnt the product, quotient and chain rules, as well as all of the minor rules (e.g. d/dx [ln(x)] = 1/x or d/dx [tan^-1] = 1/(1+x^2)), what's the next thing to learn? Surely there's more to it, I just can't find 'what's next' I guess.. ?
 

U can learn some further maths equations :')

hehehe

[lzxnl]

Hey,
So once you've learnt the product, quotient and chain rules, as well as all of the minor rules (e.g. d/dx [ln(x)] = 1/x or d/dx [tan^-1] = 1/(1+x^2)), what's the next thing to learn? Surely there's more to it, I just can't find 'what's next' I guess.. ?
 

Do you know what a derivative means? Can you apply it to other contexts? Are you comfortable with every possible application of differentiation, from finding tangents and normals to relating rates of change?

If you get bored with that, differentiate (sin x)^(sin x) and get back to me

If you really want to learn more calculus, integration is generally far harder than differentiation. Proving that the integral of sin x / x = pi/2 if integrated from 0 to infinity takes some level of extreme ingenuity in at least one step. Then there's the host of substitutions to learn, like trig subs, Weierstrauss subs, integrating by parts, integrals of quadratic radicals etc

If you get done with that, and you're done with the VCE maths courses, look up some other area of maths that interests you. For instance, linear algebra, group theory, multivariate calculus, differential equations, real analysis...there's so much out there to learn.

(For the interested people, you can do the above integral via a complex integral, differentiation under the integral sign or even interchanging the order of integration).

[TheAspiringDoc]

Do you know what a derivative means? Can you apply it to other contexts? Are you comfortable with every possible application of differentiation, from finding tangents and normals to relating rates of change?

If you get bored with that, differentiate (sin x)^(sin x) and get back to me

If you really want to learn more calculus, integration is generally far harder than differentiation. Proving that the integral of sin x / x = pi/2 if integrated from 0 to infinity takes some level of extreme ingenuity in at least one step. Then there's the host of substitutions to learn, like trig subs, Weierstrauss subs, integrating by parts, integrals of quadratic radicals etc

If you get done with that, and you're done with the VCE maths courses, look up some other area of maths that interests you. For instance, linear algebra, group theory, multivariate calculus, differential equations, real analysis...there's so much out there to learn.

(For the interested people, you can do the above integral via a complex integral, differentiation under the integral sign or even interchanging the order of integration).

Thanks lzxnl - you never cease to inspire me
I guess I think of the derivative as the slope of a curve at point x, produced by the a particular function of  x.
Finding tangents and normals hey?
So to find a tangent I believe you just plug in your x value and your good, right?
As for finding the normal, if it has the same meaning as it does in physics, then I guess you want to find the tangent, and then somehow perform a 90 degree rotation (how?).
And yeah, integration is a good idea. Since I learnt quite a bit of my first calculus from my dad, I actually ended up learning integration first as he thought it is "more intuitive". The reason I've gone back to differentiation is that in pretty much all of the integration courses, they're like "as we know from our study of derivatives" - nope.
And sorry, this probably isn't the best way of going about it, but I legit am struggling to find any info anywhere on what a differential equation is..?
Thanks again!

[MightyBeh]

So to find a tangent I believe you just plug in your x value and your good, right?

To find a tangent, you need a point from your original function or an x value; take, for example, f(x) = x²+1:

We know that the derivative is 2x,


The tangent is different at different points (mostly  ), so we need to specify a point or an x value. Let's say we have to find the tangent at x=2 - first, we have to plug this into f(x).

, which we will need to know in just a minute.

Next, we put this same x-value into the derivative:
, which we know is the slope at that point. The tangent line also lies on this point, so we can find the equation in the form y = mx + c by simply plugging in what we know:






tl;dr you also need to take the + c into account.

As for finding the normal, if it has the same meaning as it does in physics, then I guess you want to find the tangent, and then somehow perform a 90 degree rotation (how?).

Not sure what it means in Physics, but the normal is a 90 degree rotation. Not sure if it applies in 100% of cases because it's been taken off the study design for methods and I haven't covered it in class, but the slope is , where m is the gradient at a point. Again, you can sub that point in to find the + c.

And sorry, this probably isn't the best way of going about it, but I legit am struggling to find any info anywhere on what a differential equation is..?

Paul's notes can probably explain better than I can

[zsteve]

Thanks lzxnl - you never cease to inspire me
I guess I think of the derivative as the slope of a curve at point x, produced by the a particular function of  x.
Finding tangents and normals hey?
So to find a tangent I believe you just plug in your x value and your good, right?
As for finding the normal, if it has the same meaning as it does in physics, then I guess you want to find the tangent, and then somehow perform a 90 degree rotation (how?).
And yeah, integration is a good idea. Since I learnt quite a bit of my first calculus from my dad, I actually ended up learning integration first as he thought it is "more intuitive". The reason I've gone back to differentiation is that in pretty much all of the integration courses, they're like "as we know from our study of derivatives" - nope.
And sorry, this probably isn't the best way of going about it, but I legit am struggling to find any info anywhere on what a differential equation is..?
Thanks again!

A differential equation simply put is an equation that involves one or more derivatives. Examples include:
simply integrate: , this is the kind of thing you solve in SM, too easy
Other (more interesting) equations include separable, homogeneous, linear (all these are first order), second order linear homogenous/inhomogeneous equations - google for this, and Paul's notes were a great resource when I learned them. If you're as crazy as me, you would learn these in Year 12 too. (Or maybe before, seeing you're so keen. I only wish I was as diligent a learner at your age )

[zsteve]

How to differentiate . Disclaimer: I can't take any credit for sussing this out, it's something I learned a while back from a non-VCE calculus textbook (Singapore Mathematics Curriculum in fact). The method is called logarithmic differentiation
Let

and the rest follows. Sub in y when you're done.
Right |zxn|?

[lzxnl]

Yep, that works for me.
Another way is to write sin x ^ (sin x) as e^(sin x ln sin x) but that's pretty much what you've done.

[TheAspiringDoc]

Hi,
So lzxnl called finding the normal of a curve an "application" of differentiation. But how called such an application ever be applied/of use in real life?
Thanks!

[lzxnl]

Hi,
So lzxnl called finding the normal of a curve an "application" of differentiation. But how called such an application ever be applied/of use in real life?
Thanks!

If you want applications in real maths, VCE maths isn't one of the best places to find it.
Normals to surfaces, not curves, come up all the frigging time in physics. In fact, the surface normal vector is the only real way of defining the direction of a surface (think about it).
Normals to curves do come up in the two-dimensional divergence theorem which can be useful when computing double integrals (these have countless applications but again, you won't see it in VCE).

tl;dr, VCE maths sucks and if you want applications, do more maths

[GeniDoi]

If you want applications in real maths, VCE maths isn't one of the best places to find it.
Normals to surfaces, not curves, come up all the frigging time in physics. In fact, the surface normal vector is the only real way of defining the direction of a surface (think about it).
Normals to curves do come up in the two-dimensional divergence theorem which can be useful when computing double integrals (these have countless applications but again, you won't see it in VCE).

tl;dr, VCE maths sucks and if you want applications, do more maths

The probability unit of methods 3/4 is pretty applicable, and spesh is getting some probability from next year.