Apply the quotient rule, (d)/(dx)((u)/(v)) = (u'v - uv')/(v^2), where u = x^2 and v = x+1.

You're on a roll — Here are the solutions for the differentiation problems: 1. $\frac{d}{dx}\left(\frac{x^2}{x+1}\right)$ Step 1: Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = x^2$ and $v = x+1$. Step 2: Find the derivatives of $u$ and $v$. $u' = \frac{d}{dx}(x^2) = 2x$ $v' = \frac{d}{dx}(x+1) = 1$ Step 3: Substitute $u, v, u', v'$ into the quotient rule formula. $$ \frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2} $$ Step 4: Simplify the expression. $$ \frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{2x^2 + 2x - x^2}{(x+1)^2} $$ $$ \frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{x^2 + 2x}{(x+1)^2} $$ The derivative is $\boxed{\frac{x^2 + 2x}{(x+1)^2}}$. 2. $\frac{d}{dx}\left(\frac{\sin x}{x}\right)$ Step 1: Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = \sin x$ and $v = x$. Step 2: Find the derivatives of $u$ and $v$. $u' = \frac{d}{dx}(\sin x) = \cos x$ $v' = \frac{d}{dx}(x) = 1$ Step 3: Substitute $u, v, u', v'$ into the quotient rule formula. $$ \frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{(\cos x)(x) - (\sin x)(1)}{x^2} $$ Step 4: Simplify the expression. $$ \frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{x \cos x - \sin x}{x^2} $$ The derivative is $\boxed{\frac{x \cos x - \sin x}{x^2}}$. 3. $\frac{d}{dx}\left(\frac{x^3}{x^2+1}\right)$ Step 1: Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = x^3$ and $v = x^2+1$. Step 2: Find the derivatives of $u$ and $v$. $u' = \frac{d}{dx}(x^3) = 3x^2$ $v' = \frac{d}{dx}(x^2+1) = 2x$ Step 3: Substitute $u, v, u', v'$ into the quotient rule formula. $$ \frac{d}{dx}\left(\frac{x^3}{x^2+1}\right) = \frac{(3x^2)(x^2+1) - (x^3)(2x)}{(x^2+1)^2} $$ Step 4: Simplify the expression. $$ \frac{d}{dx}\left(\frac{x^3}{x^2+1}\right) = \frac{3x^4 + 3x^2 - 2x^4}{(x^2+1)^2} $$ $$ \frac{d}{dx}\left(\frac{x^3}{x^2+1}\right) = \frac{x^4 + 3x^2}{(x^2+1)^2} $$ The derivative is $\boxed{\frac{x^4 + 3x^2}{(x^2+1)^2}}$. 4. $\frac{d}{dx}\left(\frac{\ln x}{x}\right)$ Step 1: Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = \ln x$ and $v = x$. Step 2: Find the derivatives of $u$ and $v$. $u' = \frac{d}{dx}(\ln x) = \frac{1}{x}$ $v' = \frac{d}{dx}(x) = 1$ Step 3: Substitute $u, v, u', v'$ into the quotient rule formula. $$ \frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{\left(\frac{1}{x}\right)(x) - (\ln x)(1)}{x^2} $$ Step 4: Simplify the expression. $$ \frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{1 - \ln x}{x^2} $$ The derivative is $\boxed{\frac{1 - \ln x}{x^2}}$. Got more? Send 'em 📸