How do you differentiate # g(x) = 1/sqrtarctan(x^2-1)

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Question

How do you differentiate # g(x) = 1/sqrtarctan(x^2-1) 
 
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Ezekiel Alexander · Accepted Answer

#g'(x) = (-x)/(arctan^(3/2)(x^2-1)((x^2-1)^2+1)#

#g(x) = 1/(sqrt((arctan(x^2-1))#

#= [arctan(x^2-1)]^(-1/2)#

Apply power rule and chain rule

#g'(x) -1/2[arctan(x^2-1)]^(-3/2) * d/dx arctan(x^2-1)#

Apply standard derivative and chain rule

#g'(x) -1/2[arctan(x^2-1)]^(-3/2) * 1/(((x^2-1)^2+1)) * d/dx (x^2-1)#

#= -1/2[arctan(x^2-1)]^(-3/2) * 1/(((x^2-1)^2+1)) * 2x#

#= (-x)/[arctan^(3/2)(x^2-1)((x^2-1)^2+1)] #

Paisley Johnson · Answer

undefined To differentiate $g(x) = \frac{1}{\sqrt{\arctan(x^2 - 1)}}$, we use the chain rule and the derivative of $\arctan(u)$, which is $\frac{1}{1+u^2}$.

Let $u = x^2 - 1$. Then, $\frac{d}{dx}(u) = 2x$.

Now, applying the chain rule, we get:

$$\frac{d}{dx}\left(\sqrt{\arctan(u)}\right) = \frac{1}{2\sqrt{\arctan(u)}} \cdot \frac{d}{dx}\left(\arctan(u)\right)$$

$$= \frac{1}{2\sqrt{\arctan(u)}} \cdot \frac{1}{1+u^2} \cdot \frac{d}{dx}(u)$$

Substituting back $u = x^2 - 1$, we have:

$$= \frac{1}{2\sqrt{\arctan(x^2 - 1)}} \cdot \frac{1}{1+(x^2 - 1)^2} \cdot 2x$$

Now, we can simplify and rewrite the expression:

$$= \frac{x}{\sqrt{\left(1+(x^2 - 1)^2\right)\arctan(x^2 - 1)}}$$

So, the derivative of $g(x)$ with respect to $x$ is:

$$g'(x) = \frac{x}{\sqrt{\left(1+(x^2 - 1)^2\right)\arctan(x^2 - 1)}}$$

How do you differentiate # g(x) = 1/sqrtarctan(x^2-1) #?