Question

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

What is the derivative of the function?

Answer 1

Implicit differentiation is the best choice here. Take cosine of both sides to get:

\[\cos(f)=x^2+1\]
Then differentiate with respect to x.
\[-\sin (f) * f' = 2x\]
Now you can figure out what \(\sin f\) is from \(\cos f\) by remembering the pythagorean identity:
\[\cos^2 f + \sin^2 f = 1\]\[\sin f = \sqrt{1-\cos^2 f} = \sqrt{1-(x^2+1)^2}\]
Plug in and you're done!

Answer 2

I´m not sure I follow your answer

Answer 3

This is really not my area of expertise

Answer 4

First of all, lets set the question up:
\[f(x) = acos(x^2+1) \]\[let \; t=x^2+1 \]\[\therefore\ f(x) =acos(t)\]

The rule for differentiating acos is:
\[{\frac{ d }{ dx } \left[acos(x)\right] = \frac{ -1 }{ \sqrt{1-x^2}}}\]

But this is a chain rule question. The chain rule is:
\[\frac{df}{dx} = \frac{df}{dt} \times \frac{dt}{dx} \]

Substituting into the chain rule gives:
\[ \frac{df}{dx} = \frac{ -1 }{ \sqrt{1-t^2}} \times 2x \]

Substituting t into the equation and also simplifying the answer gives us:
\[ \frac{df}{dt} = \frac{2x}{\sqrt{1 - (x^2 + 1)^2}} \]

Answer 5

Could you try explaining using the chain rule (f o g)´(x)=f´(g(x))*g´(x) if it´s not too much trouble?

Answer 6

I really appreciate your help, but I´ve been using this instead. Well really it´s just (f o g)´(x) right?

Answer 7

The two functions are:
\[f(x) = acos(x) \]\[ g(x) = x^2 + 1\]

The differentials are:
\[f'(x) = \frac{-1}{\sqrt{1-x^2}}\]\[g'(x) = 2x\]

Using the chain rule:
\[ (f \circ g)'(x) = f'(g(x)) \times g'(x) \]

Then substituting in f and g:
\[ (f \circ g)'(x) = \frac{-1}{\sqrt{1-(x^2+1)^2}} \times 2x\]

Therefore the final answer is:
\[ (f \circ g)'(x) = \frac{-2x}{\sqrt{1-(x^2+1)^2}}\]

Answer 8

Thank you so much!

Answer 9

There seems to be a fatal flaw here that really bothered me last night.

Answer 10

Specifically when we get to the end, we have the square root of a negative number. So I got to thinking, OK what's the domain of \(\arccos\)? It's \([-1,1]\). What's the range of \(x^2+1\)? It's \([1,\infty)\).

So the only point this "function" works for is \(x=0\).

Answer 11

I believe you're right. But the derivative of arccos(x^2+1) is still the earlier mentioned answer? Was the domain really the issue here?

Answer 12

It's the derivative of nothing though. The answer comes out to be nonsense...

\[f'=\frac{-2}{\sqrt{-x^2-2}}\]

Answer 13

f'(x) = (-2x)/√(-x^4 - 2x^2)