Two questions - (a) Differentiating inverse trig function and (b) vector

[andrew.c]

Struggling with these two questions, any ideas?


A

Homework Statement

Differentiate with respect to x

sin^-1 2x - 4 cos^-1\frac{x}{2}

Homework Equations

The equations I have in my notes are identical to those derived here...
http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions

The Attempt at a Solution

I got

\frac{1}{1 - 2x} + \frac{8}{2 - x}

but I am totally unconvinced by myself!




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B

Homework Statement

Resolve the vector x= (1,4,-3) along and perpendicular to a = (-2,3,1).

Homework Equations

?

The Attempt at a Solution

I don't understand what the question wants me to do...

 

[Chaos2009]

For the first part, arcsin(2x), you got the derivative a little wrong.
The derivative of arcsin(x) is \frac{1}{\sqrt{1 - x^{2}}}, but x is actually 2x in this case so replace the x in the denominator of the derivative with 2x, don't forget to square it, and don't forget the chain rule! The derivative of arccos(x) is the opposite of the derivative of arcsin(x), \frac{-1}{\sqrt{1 - x^{2}}}. You can multiply the -4 through after you find the derivative of arccos(x/2), but don't forget the chain rule here either.

 

[HallsofIvy]

I strongly suspect that, using the rules you refer to, you got
\frac{1}{\sqrt{1- 4x^2}}[/itex]<br /> and reduce that to 1/(1-2x).<br /> <br /> The first is correct, the second is wrong: \sqrt{a^2- b^2} is NOT equal to a- b!

 

[Chaos2009]

\sqrt{1 - 4x^{2}} \neq 1 - 2x You just said \sqrt{a^{2} - b^{2}} \neq a - b but you went ahead and did it anyway for the first part of the problem.

\frac{d}{dx}(arcsin(2x)) = \frac{1}{\sqrt{1 - (2x)^{2}}} * \frac{d}{dx}(2x) = \frac{2}{\sqrt{1 - 4x^{2}}}
That is as reduced as it goes unless you want to rationalize the denominator which is unnecessary.

 

[timscully]

For the second question, you need to break the vector down into projection and perpendicular components.

You need to project the vector x on vector a, then take the projection of a away from x. You are then left with the perpendicular component (the same as resolving x and y components of a force vector).

The projection is given by \frac{\vec{x}.\vec{a}}{\left|\vec{a}\right|^{2}}.\vec{a}

which gives:

\vec{x}.\vec{a} = (1*-2) + (4*3) + (-3*1) = 7
\left|\vec{a}\right|^{2} = ((-2)^{2}+3^{2}+1^{2}) = 14
\frac{\vec{x}.\vec{a}}{\left|\vec{a}\right|^{2}}.\vec{a} = (7/14)*(-2,3,1) = (1/2).(-2,3,1)

Then take that away from x to get the perpendicular:

(1,4,-3) - (1/2)*(-2,3,1) = (1/2) * (4,5,-7)

The full vector is then:

x = (1/2) * (-2,3,1) + (1/2) * (4,5,-7)

Have a look at this:

http://webalg.math.tamu.edu/s03vectors/svec0601.pdf

 

[andrew.c]

Thanks for the help, I eventually did get that last answer, but I did rationalise the denominator.

Ta muchly