Let's solve these problems. Problem 6: Find the derivative of $\sin(x+y) = y^2 \cos x$. We use implicit differentiation with respect to $x$. Step 1: Differentiate both sides of the equation. $$\frac{d}{dx}(\sin(x+y)) = \frac{d}{dx}(y^2 \cos x)$$ Step 2: Apply the chain rule on the left side and the product rule on the right side. $$\cos(x+y) \cdot \frac{d}{dx}(x+y) = \frac{d}{dx}(y^2) \cdot \cos x + y^2 \cdot \frac{d}{dx}(\cos x)$$ $$\cos(x+y) \left(1 + \frac{dy}{dx}\right) = 2y \frac{dy}{dx} \cos x + y^2 (-\sin x)$$ $$\cos(x+y) + \cos(x+y) \frac{dy}{dx} = 2y \cos x \frac{dy}{dx} - y^2 \sin x$$ Step 3: Group terms containing $\frac{dy}{dx}$ on one side and other terms on the other side. $$\cos(x+y) \frac{dy}{dx} - 2y \cos x \frac{dy}{dx} = -y^2 \sin x - \cos(x+y)$$ Step 4: Factor out $\frac{dy}{dx}$ and solve for it. $$\frac{dy}{dx} (\cos(x+y) - 2y \cos x) = -(y^2 \sin x + \cos(x+y))$$ $$\boxed{\frac{dy}{dx} = \frac{-(y^2 \sin x + \cos(x+y))}{\cos(x+y) - 2y \cos x} = \frac{y^2 \sin x + \cos(x+y)}{2y \cos x - \cos(x+y)}}$$ Problem 7: Find the derivative of $f(x) = \arctan\left(\frac{2x}{1-x^2}\right)$. We can use a trigonometric substitution to simplify the function before differentiating. Step 1: Let $x = \tan\theta$. Then $\theta = \arctan x$. $$f(x) = \arctan\left(\frac{2\tan\theta}{1-\tan^2\theta}\right)$$ Step 2: Use the double angle identity $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$. $$f(x) = \arctan(\tan(2\theta))$$ $$f(x) = 2\theta$$ Step 3: Substitute $\theta = \arctan x$ back into the function. $$f(x) = 2 \arctan x$$ Step 4: Differentiate $f(x)$ with respect to $x$. $$f'(x) = \frac{d}{dx}(2 \arctan x)$$ $$f'(x) = 2 \cdot \frac{1}{1+x^2}$$ $$\boxed{f'(x) = \frac{2}{1+x^2}}$$ Problem 8: Find the derivative of $y = \ln(\sinh x + \cosh x)$. Step 1: Simplify the expression inside the logarithm. Recall that $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$. $$\sinh x + \cosh x = \frac{e^x - e^{-x}}{2} + \frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x} + e^x + e^{-x}}{2} = \frac{2e^x}{2} = e^x$$ Step 2: Substitute the simplified expression back into the function. $$y = \ln(e^x)$$ $$y = x$$ Step 3: Differentiate $y$ with respect to $x$. $$\frac{dy}{dx} = \frac{d}{dx}(x)$$ $$\boxed{\frac{dy}{dx} = 1}$$ Problem 9: Find the derivative of $y = e^{e^x}$. We use the chain rule. The derivative of $e^u$ is $e^u \frac{du}{dx}$. Step 1: Identify the outer and inner functions. Let $u = e^x$. Step 2: Differentiate $u$ with respect to $x$. $$\frac{du}{dx} = \frac{d}{dx}(e^x) = e^x$$ Step 3: Apply the chain rule. $$\frac{dy}{dx} = e^u \cdot \frac{du}{dx}$$ $$\frac{dy}{dx} = e^{e^x} \cdot e^x$$ $$\boxed{\frac{dy}{dx} = e^{x+e^x}}$$ Problem 10: Find the second derivative of $y = \frac{x}{x^2 + 1}$. Step 1: Find the first derivative $\frac{dy}{dx}$ using the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$. Let $u = x$ and $v = x^2 + 1$. Then $u' = 1$ and $v' = 2x$. $$\frac{dy}{dx} = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2}$$ $$\frac{dy}{dx} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2}$$ $$\frac{dy}{dx} = \frac{1 - x^2}{(x^2 + 1)^2}$$ Step 2: Find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{1 - x^2}{(x^2 + 1)^2}$ using the quotient rule again. Let $p = 1 - x^2$ and $q = (x^2 + 1)^2$. Then $p' = -2x$. To find $q'$, use the chain rule: $q' = 2(x^2 + 1) \cdot \frac{d}{dx}(x^2 + 1)