Can someone show me what the notation would look like in LaTex?

Original post by Apologetic Cube

Can someone show me the correct notation for a couple of questions by writing the solutions in LaTex, please?

1. Question - Differentiate e^(sin^2(x))

= e^(sin^2(x))
= e^(sin^2x) x (sin^2(x))
= e^(sin^2(x)) x (2sinx) x (cosx)
= sin2x e^(sin^2x)

2. Question - Find the equation of the tangent to the curve with the equation y = (x lnx) at the point where x = e.

Answer:

y = x lnx
y' = x(1/x) + lnx
y' = 1 + lnx

x = e

y = e lne = e

y' = 1 + lne = 2

line through (e,e) with m = 2

y - e = 2(x-e)
y = 2x - e

I hope this helps you out somewhat :smile:

1.

\begin{aligned} \dfrac{d}{dx} \left( e^{\sin^2x} \right) & = e^{\sin^2x} \times \dfrac{d}{dx} \left( \sin^2x \right) \\ & = e^{\sin^2(x)} \times 2\sin(x)\cos(x) \\ & = \sin(2x) e^{\sin^2(x)} \end{aligned}

2.

y = x \ln(x)

\begin{aligned} y' & = x \dfrac{d}{dx} \left( \ln x \right) + \ln x \dfrac{d}{dx} \left( x \right) \\ & = x \left( \dfrac{1}{x} \right) + \ln(x) \\ & = 1 + \ln(x) \end{aligned}

x = e \Rightarrow y = e\ln(e) = e

x = e \Rightarrow y' = 1 + \ln(e) = 2

\therefore y - e = 2(x-e)

y = 2x -e

(edited 10 years ago)

Reply 2

Original post by Apologetic Cube

Thanks a lot!

Could you possibly show me the working in between these two questions to arrive at the answer? The past paper book is just stating answers...

It's:

1. y = (e^x)(sinx)
dy/dx = e^x(sinx+cosx)

2. y = ln(tanx)
dy/dx = sec^2x/tanx

How do I arrive at those answers? Can you show me the working so I can follow please? Thanks a lot!

1) Use the product rule: u = e^x, v = sinx. What do you get when you apply the product rule?

2) Use the chain rule. Let u = tanx, then you have y = ln(u). Chain rule says that

dy/dx = dy/du * du/dx

What's dy/du? What's du/dx? So then what's dy/dx?

Reply 3

Original post by Apologetic Cube

However, I learned chain rule quite simplistically so I have to look into Leibniz notation for that. I'm trying:

y = ln(tanx)
dy/dx = ln(tanx) . sex^2(x)

However, I'm not sure where to go from there?

You haven't really learned chain rule, you're just trying something that seems/feels right. Chain rule can be done very quickly without substituting stuff, but trust me it's a LOT easier if you do use substitutions. Think of this as the DEFINITION of the chain rule (teachers explain chain rule very poorly in my experience):

\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}

(notice how the 'du' terms kind of cancel on the right hand side so it's easy to memorise).

Try it this way, trust me.

You know y = ln(u), so what's dy/du?

You know u = tanx, so what's du/dx?

So what's dy/dx by the chain rule?

(edited 10 years ago)

Reply 4

Original post by Apologetic Cube

I'm following you with 1) as I did:

g(x) = e^x
g'(x) = e^x

f(x) = sinx
f'(x) = cosx

g'f +f'g = e^x(sinx) + e^x(cosx)

Factor out e^x:

= e^x(sinx+cosx)

However, I learned chain rule quite simplistically so I have to look into Leibniz notation for that. I'm trying:

y = ln(tanx)
dy/dx = ln(tanx) . sex^2(x)

However, I'm not sure where to go from there?

If \ y = \ln(f(x)) \ then \ \dfrac{dy}{dx} = \dfrac{1}{f(x)} \times \dfrac{d}{dx} \left( f(x) \right)

\therefore \dfrac{d}{dx} \ln(f(x)) = \dfrac{f'(x)}{f(x)}

Reply 5

Original post by Apologetic Cube

So what would the dx be?

dy/dx = (tanx)/ln(u) x (tanx)/dx)

I have no idea how you've got that?

Forget the original problem and just separately answer these 2 questions:

1) y = ln(u), so what's dy/du?

2) u = tanx, so what's du/dx?

(imagine these were 2 completely independent and unrelated problems)

Reply 6

Original post by Apologetic Cube

I'd just be guessing if I attempted to answer that... I don't know how to find dy or du, I don't even know their relationship to y = ln(u)...

If I told you y = ln(x), you would tell me readily that dy/dx = 1/x, yup?

When I tell you y = ln(u), you should tell me that dy/du = 1/u (all I've done is change the notation from x to u, but that doesn't change calculus, it's just a letter).

Right? Ok so that answers my first question.

Now u = tanx, so what's du/dx?

(edited 10 years ago)

Reply 7

Original post by Apologetic Cube

sec² x

I really need to read into the mechanics of differentiation. At Higher (AS Level), we were only ever taught how to do differentiation for exams, we were never taught why it actually works so the notation is a little confusing for me.

Ok so we're nearly there. We worked out

dy/du = 1/u

du/dx = sec^2x.

Now chain rule tells me dy/dx = dy/du * du/dx

So

\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{1}{u}sec^2x = \frac{sec^2x}{u} = \frac{sec^2x}{tanx}

Where did the last bit come from? Well, we said u = tanx at the start.

Try these examples:

1) y = sin(cos(x))

2) y = ln(sinx + e^(x))

3) y = tan(x^3)

They all work by letting u = something nice.

(harder 4th example but still doable:

\displaystyle y = e^{e^{e^{e^x}}}

- once you can do this, you can do any chain rule question, Highers/A-level/university/etc)

(edited 10 years ago)

Reply 8

Original post by Swayum

(harder 4th example but still doable:

\displaystyle y = e^{e^{e^{e^x}}}

- once you can do this, you can do any chain rule question, Highers/A-level/university/etc)

Is this answer to example 4 ?

Unparseable latex formula:

\dfrac{d}{dx} \left( e^{e^{e^{e^x}}} \right) = e^{\left( {e^{e^{e^x}}} + {e^{e^x}}} + e^{x} + x \right)}

** That's a bit of a stretch right? Surely there are very hard uni level questions to do with the chain rule?

(edited 10 years ago)

Reply 9

Original post by Khallil

Is this answer to example 4 ?

Unparseable latex formula:

\dfrac{d}{dx} \left( e^{e^{e^{e^x}}} \right) = e^{\left( {e^{e^{e^x}}} + {e^{e^x}}} + e^{x} + x \right)}

** That's a bit of a stretch right? Surely there are very hard uni level questions to do with the chain rule?

Not quite, start by letting u = e^x, then you get

Unparseable latex formula:

e^{e^{e^u}}}

.

What can you do next?

Nope, not a stretch, once you understand dy/dx = dy/du * du/dx, you're done. At university the chain rule just extends to multiple variables, i.e.

z = sin(e^(y) + x^2) - tan(x^y)

The idea, even with multiple variables, is still virtually the same as with dy/dx = dy/du * du/dx, but the equation just gets a few more terms!

Reply 10

BlueSam3

17

Original post by Khallil

Is this answer to example 4 ?

Unparseable latex formula:

\dfrac{d}{dx} \left( e^{e^{e^{e^x}}} \right) = e^{\left( {e^{e^{e^x}}} + {e^{e^x}}} + e^{x} + x \right)}

** That's a bit of a stretch right? Surely there are very hard uni level questions to do with the chain rule?

No. That isn't how university maths works. Unless you count properly proving the chain rule, that is.

Reply 11

Original post by Swayum

Not quite, start by letting u = e^x, then you get

Unparseable latex formula:

e^{e^{e^u}}}

.

What can you do next?

Nope, not a stretch, once you understand dy/dx = dy/du * du/dx, you're done. At university the chain rule just extends to multiple variables, i.e.

z = sin(e^(y) + x^2) - tan(x^y)

The idea, even with multiple variables, is still virtually the same as with dy/dx = dy/du * du/dx, but the equation just gets a few more terms!

I just checked it through Wolfram and my answer looks like it is the right answer. The brackets enclose the power. However, my input might've been wrong or Wolfram is on the fritz :tongue:

Here is my working for your substitution :smile:

Let

u = e^x \Rightarrow \dfrac{du}{dx} = e^x

\begin{aligned} \dfrac{d}{dx} \left( e^{e^{e^{e^x}}} \right) & = \dfrac{d}{dx} \left( e^{e^{e^{u}}} \right) \\ & = e^{e^{e^{u}}} \times \dfrac{d}{dx} \left( e^{e^{u}} \right) \\ & = e^{e^{e^{u}}} \times e^{e^{u}} \times \dfrac{d}{dx} \left( e^{u} \right) \\ & = e^{e^{e^{u}}} \times e^{e^{u}} \times e^{u} \times \dfrac{du}{dx} \\ & = e^{e^{e^{e^x}}} \times e^{e^{e^x}} \times e^{e^x} \times e^x \\ & = e^{\left( e^{e^{e^x}} + e^{e^x} + e^x + x \right)} \end{aligned}

Original post by BlueSam3

No. That isn't how university maths works. Unless you count properly proving the chain rule, that is.

I'd definitely count that. Using the chain rule is hard enough as it is :lol:

(edited 10 years ago)

Reply 12

Original post by Khallil

I just checked it through Wolfram and my answer looks like it is the right answer. The brackets enclose the power. However, my input might've been wrong or Wolfram is on the fritz :tongue:

Here is my working for your substitution :smile:

Let

u = e^x \Rightarrow \dfrac{du}{dx} = e^x

\begin{aligned} \dfrac{d}{dx} \left( e^{e^{e^{e^x}}} \right) & = \dfrac{d}{dx} \left( e^{e^{e^{u}}} \right) \\ & = e^{e^{e^{u}}} \times \dfrac{d}{dx} \left( e^{e^{u}} \right) \\ & = e^{e^{e^{u}}} \times e^{e^{u}} \times \dfrac{d}{dx} \left( e^{u} \right) \\ & = e^{e^{e^{u}}} \times e^{e^{u}} \times e^{u} \times \dfrac{du}{dx} \\ & = e^{e^{e^{e^x}}} \times e^{e^{e^x}} \times e^{e^x} \times e^x \\ & = e^{\left( e^{e^{e^x}} + e^{e^x} + e^x + x \right)} \end{aligned}

I'd definitely count that. Using the chain rule is hard enough as it is :lol:

Yup that's right, sorry, the weird factorisation in the last line threw me off! I just remember it as the line just before your last one.

Reply 13