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Are you serious about needing help on g(x)? Take a good close look at it. You'll kick yourself.
These are just Chain Rule exercises.
For the first, you need to know these derivatives:
Show us what you get.
yeah i already did the g(x), its just pi because sin and arcsin cancel out and the derivative of pi(x) is just pi cause pi is constant and d/dx is just 1Originally Posted by TKHunny
Are you serious about needing help on g(x)? Take a good close look at it. You'll kick yourself.
These are just Chain Rule exercises.
For the first, you need to know these derivatives:
Show us what you get.
d/dx(1/t) is 1/t^2, d/dx(arctan(t) is 1/(1+t^2), and d/dx(3t) is 3...now what do i do from here...
Hello, mathaction!
I must assume you know the derivative of the arctangent.
. .
We have: .\:=\:\frac{1}{\arctan(3t)})
Chain Rule: .
We have: .![g(x)\;=\;\sin\left[\arcsin(\pi x)\right]](http://latex.codecogs.com/png.latex?g(x)\;=\;\sin\left[\arcsin(\pi x)\right])
Therefore: .
This one has a surprising answer . . . \;=\;\arctan(x) + \arctan\left(\frac{1}{x}\right))
We have: .
. .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
How come it works out to zero?
Let
Then: .
is in a right triangle with:
Here is the triangle:Code:* * * * * * * x * * * α * * * * * * * * * * * 1
Let
Then: .
is in a right triangle with:
Here is the triangle:Code:* * β* * * * * x * * * * * * * * * * * * * * 1
Get it?
They are complementary: .
We just differentiated a constant function: .
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