What is the derivative of $\arccos(x^2)$?
So I know that the derivative of arccos is: $-dx/\sqrt{1-x^2}$
So how would I find the derivative of $\arccos(x^2)$? What does the $-dx$ mean in the above formula?
Would it just be $-2x/\sqrt{1-x^2}$ ?
$\endgroup$ 13 Answers
$\begingroup$$$\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}$$
Using the chain rule, we can evaluate $\frac{d}{dx}\arccos (x^2)$.
$$\frac{d}{dx}\arccos(x^2)=-\frac{1}{\sqrt{1-(x^2)^2}}\frac{d}{dx}(x^2)=-\frac{1}{\sqrt{1-x^4}}\frac{d}{dx}(x^2)=-\frac{1}{\sqrt{1-x^4}}\cdot 2x$$
$\endgroup$ $\begingroup$If you have a composite function $h(x)=(g\circ f)(x)$ then its derivative is given by $$h'(x)=g'(f(x))f'(x).$$ (This is known as the Chain rule. See for more information.)
In your case, $f(x)=x^2$ and $g(x)=\arccos x.$ Since $f'(x)=2x$ and $g'(x)=\frac{-1}{\sqrt{1-x^2}},$ you can get the derivative just applying the Chain rule. (Note that $g'$ must be evaluated at $f(x)$ and not at $x.$)
$\endgroup$ $\begingroup$"$-dx$" should not be there.
$$ \frac{d}{dx} \arccos(x^2) = \frac{-1}{\sqrt{1-(x^2)^2}} \cdot \frac d {dx} x^2 = \frac{-2x}{\sqrt{1-x^4}}. $$
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