Derivative

[lec87]

If f(x) = InverseCos[2x/(1+x^2)]
Find f '(x)

After some workings I got the answer f '(x) = 2/(1+x^2), but I think the solution is actually -2/(1+x^2). Does anyone know what the correct answer is?

[krtxmrtz]

I assume by InverseCos you mean the ArcCos or cos^-1. The derivative of this function can be found from:

f(x) = cos^-1(x)

x = cos(f)

df/dx = 1 / (dx/df) = 1 / -sin(f) = 1 / -sqrt(1 - cos²(f)) = -1 / sqrt(1 - x²)

Now, applying the chain rule for the composite functions:

f(x) = cos^-1[2x/(1+x²)]

df/dx = (-1 / sqrt(1 - (2x/(1 + x²)))) * d/dx (2x/(1+x²)) = -2/(1 + x²)

[sabke_papa]

Your question was -

Differentiate f(x) with respect to x where -

f(x) = cos^(-1) { 2*x/(1+x^2)}

Putting x = tan y first,f(x) becomes-

f(x) = cos^(-1) [ 2*tan y/{1+(tan y)^2} ]

f(x) = cos^(-1) { 2*tan y/(sec y)^2 }

f(x) = cos^(-1) [ 2*sin y*cos y ]

f(x) = cos^(-1) [ sin 2*y ]

f(x) = cos^(-1) [ cos (90 - 2*y) ] 90 is in degrees

f(x) = 90 - 2*y

We substituted x = tan y, so y = tan^(-1)(x) -

f(x) = 90 - tan^(-1)(x)

Now differentiate it -

d/dx[f(x)] = -2*x/(1+x^2)

HTH.

[krtxmrtz]

I made a mistake in the last line of my post above so, I've corrected it. But sabke_papa's solution is more elegant, therefore I don't need go through all the trouble of the details.
By the way,

Now differentiate it -

d/dx[f(x)] = -2*x/(1+x^2)

the *x after the -2 must be removed. But of course it's a typo.