Math / Trigonometry

Question(3) $y=\operatorname{arcCsc}\left(x^{2}+1\right) ; x>0$

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of the inverse cosecant of $x^2 + 1$ where $x$ is greater than 0. Watch out! Don't forget the chain rule, and remember the derivative of $\operatorname{arcCsc}(x)$ has an absolute value in it!
Also, remember to keep track of that $x > 0$ condition!

STEP 2

1. Rewrite with Sine
2. Differentiate
3. Simplify

STEP 3

Let's rewrite our function $y = \operatorname{arcCsc}(x^2 + 1)$ in terms of sine.
Remember that $\csc(\theta) = \frac{1}{\sin(\theta)}$ , so if we take the cosecant of both sides, we get $\csc(y) = x^2 + 1$ .
Then, we can flip both sides to get $\sin(y) = \frac{1}{x^2 + 1}$ .
This will make taking the derivative a bit easier!

STEP 4

Now, let's differentiate both sides of $\sin(y) = \frac{1}{x^2 + 1}$ with respect to $x$ .
Using the chain rule on the left side, we get $\cos(y) \cdot \frac{dy}{dx}$ .
For the right side, we can use the power rule after rewriting $\frac{1}{x^2 + 1}$ as $(x^2 + 1)^{-1}$ .
So, the derivative of $(x^2 + 1)^{-1}$ with respect to $x$ is $-1(x^2 + 1)^{-2} \cdot 2x$ , which simplifies to $\frac{-2x}{(x^2 + 1)^2}$ .

STEP 5

Putting it together, we have $\cos(y) \cdot \frac{dy}{dx} = \frac{-2x}{(x^2 + 1)^2}$ .
Now, we want to solve for $\frac{dy}{dx}$ , so we divide both sides by $\cos(y)$ to get $\frac{dy}{dx} = \frac{-2x}{\cos(y)(x^2 + 1)^2}$ .

STEP 6

We need to replace $\cos(y)$ with something in terms of $x$ .
Remember the Pythagorean identity: $\sin^2(y) + \cos^2(y) = 1$ .
We know that $\sin(y) = \frac{1}{x^2 + 1}$ , so $\sin^2(y) = \frac{1}{(x^2 + 1)^2}$ .
Substituting this into our identity, we get $\frac{1}{(x^2 + 1)^2} + \cos^2(y) = 1$ .
Solving for $\cos^2(y)$ , we have $\cos^2(y) = 1 - \frac{1}{(x^2 + 1)^2} = \frac{(x^2 + 1)^2 - 1}{(x^2 + 1)^2} = \frac{x^4 + 2x^2}{(x^2 + 1)^2}$ .
Taking the square root of both sides gives us $\cos(y) = \pm \frac{\sqrt{x^4 + 2x^2}}{x^2 + 1} = \pm \frac{x\sqrt{x^2 + 2}}{x^2 + 1}$ .

STEP 7

Since $x > 0$ and $\operatorname{arcCsc}(\theta)$ has a range of $[-\pi/2, 0) \cup (0, \pi/2]$ , $y$ must be in the first or fourth quadrant.
In these quadrants, cosine is positive, so we take the positive square root.
Thus, $\cos(y) = \frac{x\sqrt{x^2 + 2}}{x^2 + 1}$ .

STEP 8

Finally, substitute this back into our expression for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{-2x}{\left(\frac{x\sqrt{x^2 + 2}}{x^2 + 1}\right)(x^2 + 1)^2} = \frac{-2x(x^2 + 1)}{x\sqrt{x^2 + 2}(x^2 + 1)^2} = \frac{-2}{\sqrt{x^2 + 2}(x^2 + 1)}$ .

STEP 9

Our derivative is $\frac{dy}{dx} = \frac{-2}{\sqrt{x^2 + 2}(x^2 + 1)}$ .
This is as simplified as it gets!

STEP 10

The derivative of $y = \operatorname{arcCsc}(x^2 + 1)$ for $x > 0$ is $\frac{-2}{\sqrt{x^2 + 2}(x^2 + 1)}$ .

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