Derivative??????

Question

How is f(x)=Arctan 8/x^2 done?????????

Answer 1

First the derivative of arctan x = 1/(1+x^2)
But its not just x so you will need to use the chain rule and multiple by the derivative of 8/x^2.
So:
f ' (x) = 1/(1+(8/x^2)^2) * (-16/x^3)
= -16 / ((1+64/x^4)x^3)
= -16x / ((1+64/x^4)x^4)
= -16x / (x^4+64)

Answer 2

Learning this will allow you to do lots of others.

Let y = Arctan (8/(x^2)), then tan y = 8/(x^2)

Take the derivative of both sides, so dy/(cos y)^2 = -16dx/(x^3)
so dy/dx = -16*(cos y)^2/(x^3). What is y? y = Arctan (8/(x^2)), so drawing the triangle, opposite/adjacent = 8/(x^2). To get cos y, we need adjacent/hypotenuse, so
cos y = x^2/sqrt(64 + x^4). Therefore, dy/dx = -16*(x^4/(64+x^4))/x^3 = -16x/(64 + x^4).

The two tricks are to set y = f(x) and take the derivative of both sides and then to calculate cos y from trigonometry.

Answer 3

y = tan^(-1) ( 8 / x ² )
Let u = 8 / x ² = 8 x^( - 2 )
du / dx = - 16 / x ³
y = tan^(-1) u
dy/du = 1 / (1 + u ²)

dy / dx = (dy / du) (du / dx)
dy / dx = [ 1 / (1 + u ²) ] [- 16 / x ³ ]
dy / dx = [x^4 / (x ^ 4 + 64 ) ] [ - 16 / x ³ ]
dy / dx = - 16 x / ( x ^ 4 + 64 )