Differentiation of loagrithms (1 Viewer)

[cutemouse]

Of course ln2 is a real number.

It's log_e2

If x=log_e2, then it is the same as saying that e^x=2

What are you trying to say?

 

[lolokay]

the OP was confused (asking why ln2 isn't differentiated)
jetblack was just pointing out that ln2 is a number/constant, not a function, so it is treated the same as any other constant when differentiating

 

[micuzzo]

log₂x=log_ex/log_e2 [Log rules of changing base]

By quotient rule, d/dx=(vu'-uv')/v²

Therefore, d/dx=[(log_e2).(1/x)-log_ex.0]/(log_e2)²

=1/xlog_e2
=1/xln2

Hope that helps

CHAMPION... thanks heaps buddy.... i was looking at the question in the wrong way...

p.s: thanks also to 'lyounamu'

 

[bored of sc]

On the topic of differentiation, I have a question that I’m stuck with:

Differentiate: y = sin<SUP>3</SUP>x / cos<SUP>4</SUP>x

Thanx.

I haven't done this topic yet but I reckon you might have to simplify first

i.e. y = tan³x/cosx

 

[dp624]

you can use quotient or product rule if you've learnt it?

y' = (cos ⁴ x)(3 sin ² x cos x) - (sin ³ x)(-)(4 cos ³x sin x) OVER cos ⁸ x

i hope i havent forgotten my maths stuff already haha

 

[jet]

You use the quotient rule.
We know, d/dx(sin^3(x))=3cos(x)sin^2(x)
and
d/dx(cos^4(x))=-4sin(x)cos^3(x)
Hence, d/dx(sin^3(x)/cos^4(x))
= (vu'-uv')(v^20
=(3cos^5(x)sin^2(x)+4sin^4(x)cos^3(x))/(cos^8(x))
=(3cos^2(x)sin^2(x) + 4sin^4(x))/cos^5(x)

 

[Trebla]

On the topic of differentiation, I have a question that I’m stuck with:

Differentiate: y = sin<SUP>3</SUP>x / cos<SUP>4</SUP>x

Thanx.

Product rule might be easier lol:
y = sin³x / cos⁴x
= tan³x.sec x
dy/dx = 3tan²x.sec²x.sec x + tan³x.sec x.tan x
= 3tan²x.sec³x + tan⁴x.sec x

 

[jet]

The only thing with that is that 2 unit isn't required to differentiate secx. I think that this would definitely be a quotient rule question.

 

[Trebla]

The only thing with that is that 2 unit isn't required to differentiate secx. I think that this would definitely be a quotient rule question.

I'm pretty sure the derivative of sec x is within the scope of the 2 unit course. The only part of the Trigonometric Functions topic that is strictly Extension 1 according to the syllabus is the integral of cos²x and sin²x because it requires double angles, which are not in the 2 unit syllabus.

 

[jet]

No worries. What textbook?

 

[Trebla]

Thanx the answers I had must of been wrong cause i had the same thing bu the answers said

3sin<SUP>2</SUP>x + sin<SUP>4</SUP>x /cos<SUP>5</SUP>x

I think it meant (3sin<SUP>2</SUP>x + sin<SUP>4</SUP>x) /cos<SUP>5</SUP>x

 

[Timothy.Siu]

derivative of sec x shud be 2unit,...its just (cos x)^-1

 

[Trebla]

The answers are actually equivalent. They're all correct.

Note that from jetblack2007's answer:
(3sin²x.cos²x + 4sin⁴x) / cos⁵x
= (3sin²x.(1 - sin²x) + 4sin⁴x) / cos⁵x
= (3sin²x - 3sin⁴x + 4sin⁴x) / cos⁵x
= (3sin²x + sin⁴x) / cos⁵x
which is the other form of the answer
= 3sin²x / cos²x.cos³x + sin⁴x / cos⁴x.cos x
= 3tan²x / cos³x + tan⁴x / cos x
= 3tan²x.sec³x + tan⁴x.sec x
which is the another form of the answer that I had...