First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

[InteGrand]

Re: MATH1131 help thread

Yes that's what I'm asking how do I do that?

E.g. When x = 0, the value is negative. When x = 3, the value is positive.

(In other words inspection as leehuan said above.)

 

[Flop21]

Re: MATH1131 help thread

That's not even hard. Just put random numbers (for x) into your calculator and get something >0 and something <0 ?

E.g. prove your function has a root.
I could pick 0 and 10

Or I could pick -1000 and 1000

oh, thought there was some math trick way

sweet thanks

 

[leehuan]

Re: MATH1131 help thread

oh, thought there was some math trick way

sweet thanks

There are a few but...nah I just sub in random (however tiny bit cleverly chosen) values until profit

 

[Flop21]

Re: MATH1131 help thread

What's happening here,

log y = log(sinx)^x

= xlog(sinx)


So I get they added log to both sides but don't get how it turns into the second part.

 

[leehuan]

Re: MATH1131 help thread

What's happening here,

log y = log(sinx)^x

= xlog(sinx)


So I get they added log to both sides but don't get how it turns into the second part.

log(A^B) = B*log(A)

Logarithm law

 

[iforgotmyname]

Re: MATH1131 help thread

What's happening here,

log y = log(sinx)^x

= xlog(sinx)


So I get they added log to both sides but don't get how it turns into the second part.

Log rules, ln(x^Y) = y*ln(x)

are you sure the power x is outside the bracket?

 

[InteGrand]

Re: MATH1131 help thread

Log rules, ln(x^Y) = y*ln(x)

are you sure the power x is outside the bracket?

Judging by the next line, it's surely inside the log, i.e. log(sin(x)^x).

 

[Flop21]

Re: MATH1131 help thread

Log rules, ln(x^Y) = y*ln(x)

are you sure the power x is outside the bracket?

Yes it's outside. It's outside on the original function as well (before they added ln)

 

[leehuan]

Re: MATH1131 help thread

Yes it's outside. It's outside on the original function as well (before they added ln)

 

[InteGrand]

Re: MATH1131 help thread

Yes it's outside. It's outside on the original function as well (before they added ln)

How do you know it's outside? (It could just be laziness about brackets. It should be clear from the context which one was meant.)

Was the question asking to differentiate (sin x)^x or something (if so, logs would be taken and then logarithmic differentiation used, and it'd be log((sin x)^x).)

 

[Flop21]

Re: MATH1131 help thread

How do you know it's outside? (It could just be laziness about brackets. It should be clear from the context which one was meant.)

Was the question asking to differentiate (sin x)^x or something (if so, logs would be taken and then logarithmic differentiation used, and it'd be log((sin x)^x).)

Because it's literally outside brackets "( )". But okay yeah I understand thanks.

 

[InteGrand]

Re: MATH1131 help thread

Because it's literally outside brackets "( )". But okay yeah I understand thanks.

If it's written log(sin x)^x, from the context (I'm guessing), it probably means the log of: (sin x)^x.

They didn't write (log(sin x))^x, did they? If they did, then that'd definitely be different.

 

[Flop21]

Re: MATH1131 help thread

If it's written log(sin x)^x, from the context (I'm guessing), it probably means the log of: (sin x)^x.

They didn't write (log(sin x))^x, did they? If they did, then that'd definitely be different.

yes written like that

 

[leehuan]

Re: MATH1131 help thread

yes written like that

Usually that means log((sinx)^x)

Not the other one

(in reality it usually means log(sin(x^x)) but let's just ignore that case

 

[InteGrand]

Re: MATH1131 help thread

yes written like that

Yeah, then it'd definitely mean the log of: (sin x)^x (especially if it had y = (sin x)^x before that, which'd mean they're taking logs of both sides next line).

 

[Flop21]

Re: MATH1131 help thread

No I never said that at all?

I meant what I wrote. There were never any brackets around the whole log function.

log y = log(sinx)^x

That to me says

 

[Flop21]

Re: MATH1131 help thread

Integrating by parts...

(1-x)/(1+x)^3 dx

I'm confused about how to assign u and v, and thus u' and v'.

The answers have v' = 1/(1+x)^3 and I'm confused why.

 

[InteGrand]

Re: MATH1131 help thread

Integrating by parts...

(1-x)/(1+x)^3 dx

I'm confused about how to assign u and v, and thus u' and v'.

The answers have v' = 1/(1+x)^3 and I'm confused why.

Basically in integration by parts, we want to pick one factor that we'll differentiate and one that we'll integrate.

Here, we'll want to integrate the 1/(1+x)^3 and differentiate the 1 – x.

The reason for this is that when we differentiate the 1 – x, it'll just become -1, and when we integrate the 1/(1+x)^3, it's something that we'll be able to integrate easily (namely -(1/2)*(1/(1+x)^2)) and thus complete the integration.

 

[Flop21]

Re: MATH1131 help thread

Basically in integration by parts, we want to pick one factor that we'll differentiate and one that we'll integrate.

Here, we'll want to integrate the 1/(1+x)^3 and differentiate the 1 – x.

The reason for this is that when we differentiate the 1 – x, it'll just become -1, and when we integrate the 1/(1+x)^3, it's something that we'll be able to integrate easily (namely -(1/2)*(1/(1+x)^2)) and thus complete the integration.

How do we do that?

Think I'm just confused how to do these questions involving fractions you cannot simplify and have to do with integration by parts. Can do integration by parts if it's just f(x)*g(x).

 

[InteGrand]

Re: MATH1131 help thread

How do we do that?

Think I'm just confused how to do these questions involving fractions you cannot simplify and have to do with integration by parts. Can do integration by parts if it's just f(x)*g(x).