Find the derivative of the function. $$ y=3 an ^{-1}\left(x+\sqrt{1+x^{2}} ight) $$ $$ y^{\prime}=\square $$

Question

Find the derivative of the function. $$ y=3 	an ^{-1}\left(x+\sqrt{1+x^{2}}ight) $$ $$ y^{\prime}=\square $$

UpStudy AI · Accepted Answer

Let
$$
y=3\,\tan^{-1}\Bigl(x+\sqrt{1+x^2}\Bigr).
$$

We want to find $$y'$$. One way to do this is to differentiate directly; an alternative is to use a substitution that simplifies the expression. We will use a hyperbolic substitution.

**Step 1. Use a Hyperbolic Substitution**

Let
$$
x=\sinh t.
$$
Then
$$
\sqrt{1+x^2}=\sqrt{1+\sinh^2t}=\cosh t.
$$
Thus,
$$
x+\sqrt{1+x^2} = \sinh t+\cosh t = e^t.
$$
So the function becomes
$$
y=3\,\tan^{-1}\Bigl(e^t\Bigr).
$$

**Step 2. Differentiate with Respect to $$t$$**

Differentiate $$y$$ with respect to $$t$$:
$$
\frac{dy}{dt}=3\cdot\frac{d}{dt}\Bigl[\tan^{-1}\bigl(e^t\bigr)\Bigr]=3\cdot\frac{1}{1+e^{2t}}\cdot e^t.
$$
Thus,
$$
\frac{dy}{dt}=\frac{3e^t}{1+e^{2t}}.
$$

**Step 3. Express $$\frac{dy}{dx}$$ in Terms of $$t$$**

We have $$x=\sinh t$$, so
$$
\frac{dx}{dt}=\cosh t.
$$
Then, by the chain rule,
$$
\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=\frac{3e^t}{1+e^{2t}}\cdot\frac{1}{\cosh t}.
$$

Notice that
$$
\frac{e^t}{1+e^{2t}} = \frac{1}{e^{-t}+e^t}=\frac{1}{2\cosh t},
$$
since
$$
e^{-t}+e^{t}=2\cosh t.
$$
Thus,
$$
\frac{dy}{dx}=\frac{3}{2\cosh t}\cdot\frac{1}{\cosh t} = \frac{3}{2\cosh^2t}.
$$

**Step 4. Rewrite in Terms of $$x$$**

Recall that
$$
\cosh t=\sqrt{1+\sinh^2t}=\sqrt{1+x^2}.
$$
So,
$$
\cosh^2t=1+x^2.
$$
Therefore,
$$
y'=\frac{3}{2(1+x^2)}.
$$

**Final Answer**

$$
y'=\frac{3}{2(1+x^2)}.
$$
$$ y' = \frac{3}{2(1 + x^2)} $$

Find the derivative of the function. \( y=3 \tan ^{-1}\left(x+\sqrt{1+x^{2}}\right) \) \( y^{\prime}=\square \)

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