Question

f(x) = arccos(x^2+1)

What´s the derivative of the function?

Answer 1

Do you know the chain rule? Do you know the derivative of the inverse cosine function?

Answer 2

Well, the derivative of arccos is -1/√1-x^2, right?

Answer 3

\[\frac{ d }{ dx }(\cos^{-1} u)=\frac{ -1 }{ \sqrt{1-u^2} }u \prime \]
where u is a function of x

Answer 4

yes follow what the above post says.
the substitution is that
u = x^2 + 1
du = 2x

Answer 5

Using that substitution
\[u = x^2 + 1\]
\[\frac{ du }{ dx }=2x\]
the chain rule is
\[\frac{ df }{ dx }=\frac{ df }{ du }*\frac{ du }{ dx }\]

Answer 6

\[f(u)=\cos^{-1} (u)\]
\[\frac{ df }{ dx }=\frac{ df }{ du }*\frac{ du }{ dx } = \frac{ -1 }{\sqrt{ 1-u^2} }*\frac{ du }{ dx }\]

Answer 7

just replace back to f(x) for u and du/dx

Answer 8

\[\frac{ df }{ dx }=\frac{ -1 }{ \sqrt{1-(x^2 + 1)^2} } * 2x\]

Answer 9

not sure if that root can be a real number though