Calculus

1. y=sin^-1(2x)

find dy/dx

(im guessing it's the chain rule, let 2x=u, but am a bit rusty after that bit)

2.
f(x) = (sinx + 2)e^sinx

show f '(x) = e^x.cosx(sin + 3)

(I get as far as cosx.e^sinx(sinx+2) + e^sinx(cozx)...right track or barking up the wrong tree?)

3. a function is defined by f(x) = x+1 / (2x-1)^.5

find the co-ordinates at the stationary point(a minumum).

Any help much appreciated!

Reply 1

J.F.N

1

jumpunderaboat

1. y=sin^-1(2x)

find dy/dx

(im guessing it's the chain rule, let 2x=u, but am a bit rusty after that bit)

2.
f(x) = (sinx + 2)e^sinx

show f '(x) = e^x.cosx(sin + 3)

(I get as far as cosx.e^sinx(sinx+2) + e^sinx(cozx)...right track or barking up the wrong tree?)

3. a function is defined by f(x) = x+1 / (2x-1)^.5

find the co-ordinates at the stationary point(a minumum).

Any help much appreciated!

1. sin^-1 x has a standard derivative = 1/sqrt(1-x^2) (let me know if you want to know how we arrive at it), so dy/dx=1/sqrt(1-4x^2)
2. (sinx + 2)e^sinx
differentiate as a product
cosx.(sinx + 2)e^sinx + e^sinx.(cosx)
factor out cosx.e^sinx
cosx.e^sinx.(sinx+2+1)=cosx.e^sinx(sinx+3)
3. Differentiate as a quotient and equate to zero to get...
(2x-1)^5 - (x+1).10(2x-1)^4 =0
Factory out (2x-1)^4.(2x-1 +10x+10)=0
(2x-1)^4.(12x-9)=0
So we have the minima is when 12x-9=0
-->x=3/4 --> y=56

someone should check over my work on the last question since i did it hastily

Reply 2

Jump

OP

18

Thanks loads! and could I take you up on that offer as I don't think I've come across that standard derivative before. Also would it not be dy/dx=2/sqrt(1-4x^2)?

Reply 3

Newton

1. f(x)=sin^-1(2x)

If by that you mean f(x)=arcsin(2x)

Then by applying the chain rule we let f(u)=arcsin(u) where u=2x

(dy/dx)=(dy/du)*(du/dx)

(dy/du)=(1/(sqrt(1-(u^2)))

AND

(du/dx)=2

Therefore:

(dy/dx)=(2/(sqrt(1-4(x^2)))

If you mean f(x)=((sin2x)^(-1))=cosec2x

Then apply the chain rule; f(u)=cosecu where u=2x

(dy/du)=-cosecucotu

AND

(du/dx)=2

Therefore:

(dy/dx)=-2cosec2xcot2x

2.f(x)=(sinx + 2)(e^(sinx))

In this case you have to apply the product rule:

(uv)'=vu'+uv'

Let u=sinx+2=>u'=cosx

AND

Let v=(e^(sinx))=>v'=(cosx)(e^(sinx))

Therefore:

(uv)'=(e^(sinx))(cosx)+(sinx+2)(cosx)(e^(sinx))

Take the common factor of (e^(sinx))(cosx):

=>(uv)'=(e^(sinx))(cosx)(1+(sinx+2))

=>(uv)'=(e^(sinx))(cosx)(sinx+3)

3. f(x)=((x+1)/((2x-1)^5))

Apply the quotient rule:

(u/v)'=((vu'-uv')/(v^2))

Let u=x+1=>u'=1

AND

Let v=((2x-1)^5)=>v'=10((2x-1)^4)

Therefore:

(u/v)'=((((2x-1)^5)-10(x+1)((2x-1)^4)/((2x-1)^10))

Take the common factor of ((2x-1)^4) outside on the top:

=>(u/v)'=(((2x-1)^4)((2x-1)-10(x+1))/((2x-1)^10))

=>(dy/dx)=((-8x-11)/((2x-1)^6))

A turning point occurs when (dy/dx)=0

=>0=-8x-11

=>x=(-11/8)=>y=0.001

Newton.

Reply 4

Fermat

8

1.
As an alternative to the quotient rule shown, you can do it this way also,

y=sin^(-1)(2x)
2x = siny
x = ½siny
dx/dy = ½cosy
dx/dy = ½√(1 - sin²y)
dx/dy = ½√(1 - 4x²)
dy/dx = 1 ÷ dx/dy
dy/dx = 2/√(1 - 4x²)
===============

Reply 5

Newton

Fermat

1.
As an alternative to the quotient rule shown, you can do it this way also,

y=sin^(-1)(2x)
2x = siny
x = ½siny
dx/dy = ½cosy
dx/dy = ½√(1 - sin²y)
dx/dy = ½√(1 - 4x²)
dy/dx = 1 ÷ dx/dy
dy/dx = 2/√(1 - 4x²)
===============

Yes, but that is just deriving the derivative from the first principles. Since the derivative is a standard one it is much better to use the chain rule, not the quotient rule.

Newton.

Reply 6

Fermat

8

Newton

Yes, but that is just deriving the derivative from the first principles...

Newton.

Yes, that is obtaining the derivative from first principles. Which, apart from being an alternative method , is sometimes very handy to know. Especially during an exam, when you know that the standard derivative involves a quotient with a quadratic term on the bottom, inside a square root sign.
But is it √(1 - x²), √(x² - 1) or √(1 + x²) ? If your memory isn't too good, then knowing how to quickly obtain a result from 1st principles can be life-saving.
This method can be used on other trig functions to obtain standard derivatives

Newton

... Since the derivative is a standard one it is much better to use the chain rule, not the quotient rule.

Newton.

I didn't use the quotient rule.

Calculus

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