$y = \tan^{-1} \left( \frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x} \right) + \tan^{-1} \left( \frac{x}{1 + \sqrt{1 + x^2}} \right)$, then find $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$

Question

The interval in which $ y = x^2 e^x $ is decreasing is _____

Answer 1

To find the derivative $\frac{dy}{dx}$ of the given expression at $x = \frac{\sqrt{3}}{2}$, we start by analyzing the expression:

$y = \tan^{-1} \left( \frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x} \right) + \tan^{-1} \left( \frac{x}{1 + \sqrt{1 + x^2}} \right)$

This involves using the derivative of inverse tangent, particularly knowing that:

The derivative of $\tan^{-1}(u)$ is $\frac{du/dx}{1 + u^2}$.

Let's differentiate each part starting with the first term:

Let $u = \frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}$.
1. Take derivative of $u$ using the quotient rule: $\frac{du}{dx} = \frac{(4\cos x + 3\sin x)(-3\sin x - 4\cos x) - (3\cos x - 4\sin x)(-4\sin x - 3\cos x)}{(4\cos x + 3\sin x)^2}$.
2. Simplify this derivative to find $\frac{du}{dx}$.
3. Apply the inverse tangent derivative formula: $\frac{d}{dx}[\tan^{-1}(u)] = \frac{\frac{du}{dx}}{1 + u^2}$.
Let $v = \frac{x}{1 + \sqrt{1 + x^2}}$.
1. Take derivative of $v$ using the quotient rule and chain rule: $\frac{dv}{dx} = \frac{(1 + \sqrt{1+x^2}) - x \cdot \frac{x}{2\sqrt{1+x^2}}}{(1+\sqrt{1+x^2})^2}$.
2. Simplify to find $\frac{dv}{dx}$.
3. Apply the inverse tangent derivative: $\frac{d}{dx}[\tan^{-1}(v)] = \frac{\frac{dv}{dx}}{1 + v^2}$.

Combine the derivatives obtained:

$\frac{dy}{dx} = \frac{\frac{du}{dx}}{1 + u^2} + \frac{\frac{dv}{dx}}{1 + v^2}$

Finally, substitute $x = \frac{\sqrt{3}}{2}$ into the final expression to evaluate:

The calculation gives:

Correct answer: $-\frac{5}{7}$

Show Hint