Check the form of the limit. As x infinity, both the numerator (x^2 - x) and the denominator (x^3 + 2x) approach infinity. This is an indeterminate form (infinity)/(infinity), so L'Hôpital's

Alright — let's do this. 1. To evaluate $\lim_{x \to \infty} \frac{x^2 - x}{x^3 + 2x}$ using L'Hôpital's rule: Step 1: Check the form of the limit. As $x \to \infty$, both the numerator $(x^2 - x)$ and the denominator $(x^3 + 2x)$ approach $\infty$. This is an indeterminate form $\frac{\infty}{\infty}$, so L'Hôpital's rule can be applied. Step 2: Differentiate the numerator and the denominator. $$ \frac{d}{dx}(x^2 - x) = 2x - 1 $$ $$ \frac{d}{dx}(x^3 + 2x) = 3x^2 + 2 $$ Step 3: Apply L'Hôpital's rule. $$ \lim_{x \to \infty} \frac{x^2 - x}{x^3 + 2x} = \lim_{x \to \infty} \frac{2x - 1}{3x^2 + 2} $$ Again, as $x \to \infty$, this is an indeterminate form $\frac{\infty}{\infty}$. Apply L'Hôpital's rule again. Step 4: Differentiate the new numerator and denominator. $$ \frac{d}{dx}(2x - 1) = 2 $$ $$ \frac{d}{dx}(3x^2 + 2) = 6x $$ Step 5: Apply L'Hôpital's rule a second time. $$ \lim_{x \to \infty} \frac{2x - 1}{3x^2 + 2} = \lim_{x \to \infty} \frac{2}{6x} $$ Step 6: Evaluate the limit. As $x \to \infty$, $6x \to \infty$, so $\frac{2}{6x} \to 0$. $$ \lim_{x \to \infty} \frac{2}{6x} = 0 $$ The final answer is $\boxed{0}$. 2. Differentiate the following functions: a) $y = \frac{\sqrt[3]{(1+3x)^2}}{x}$ Step 1: Rewrite the function using exponents. $$ y = \frac{(1+3x)^{2/3}}{x} $$ Step 2: Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$. Let $u = (1+3x)^{2/3}$ and $v = x$. Step 3: Find $u'$ using the chain rule. $$ u' = \frac{2}{3}(1+3x)^{(2/3)-1} \cdot \frac{d}{dx}(1+3x) $$ $$ u' = \frac{2}{3}(1+3x)^{-1/3} \cdot 3 $$ $$ u' = 2(1+3x)^{-1/3} $$ Step 4: Find $v'$. $$ v' = \frac{d}{dx}(x) = 1 $$ Step 5: Substitute $u, v, u', v'$ into the quotient rule formula. $$ \frac{dy}{dx} = \frac{2(1+3x)^{-1/3} \cdot x - (1+3x)^{2/3} \cdot 1}{x^2} $$ Step 6: Simplify the expression. Factor out $(1+3x)^{-1/3}$ from the numerator. $$ \frac{dy}{dx} = \frac{(1+3x)^{-1/3} [2x - (1+3x)^{2/3}(1+3x)^{1/3}]}{x^2} $$ $$ \frac{dy}{dx} = \frac{(1+3x)^{-1/3} [2x - (1+3x)]}{x^2} $$ $$ \frac{dy}{dx} = \frac{(1+3x)^{-1/3} [2x - 1 - 3x]}{x^2} $$ $$ \frac{dy}{dx} = \frac{(1+3x)^{-1/3} [-x - 1]}{x^2} $$ $$ \frac{dy}{dx} = \frac{-(x+1)}{x^2 (1+3x)^{1/3}} $$ The final answer is $\boxed{-\frac{x+1}{x^2 \sqrt[3]{1+3x}}}$. b) $y = \csc\left(\frac{3}{4}x^2\right)$ Step 1: Apply the chain rule. The derivative of $\csc(u)$ is $-\csc(u)\cot(u) \cdot u'$. Let $u = \frac{3}{4}x^2$. Step 2: Find $u'$. $$ u' = \frac{d}{dx}\left(\frac{3}{4}x^2\right) = \frac{3}{4} \cdot 2x = \frac{3}{2}x $$ Step 3: Substitute $u$ and $u'$ into the chain rule formula. $$ \frac{dy}{dx} = -\csc\left(\frac{3}{4}x^2\right)\cot\left(\frac{3}{4}x^2\right) \cdot \left(\frac{3}{2}x\right) $$ Step 4: Rearrange the terms. $$ \frac{dy}{dx} = -\frac{3}{2}x \csc\left(\frac{3}{4}x^2\right)\cot\left(\frac{3}{4}x^2\right) $$ The final answer is $\boxed{-\frac{3}{2}x \csc\left(\frac{3}{4}x^2\right)\cot\left(\frac{3}{4}x^2\right)}$. c) $x^2y + y^2 + 4x = 1$ Step 1: Differentiate both sides of the equation with respect to $x$. Remember to use implicit differentiation for terms involving $y$. Step 2: Differentiate $x^2y$ using the product rule. $$ \frac{d}{dx}(x^2y) = \frac{d}{dx}(x^2) \cdot y + x^2 \cdot \frac{d}{dx}(y) = 2xy + x^2\frac{dy}{dx} $$ Step 3: Differentiate $y^2$ using the chain rule. $$ \frac{d}{dx}(y^2) = 2y\frac{dy}{dx} $$ Step 4: Differentiate $4x$. $$ \frac{d}{dx}(4x) = 4 $$ Step 5: Differentiate the constant $1$. $$ \frac{d}{dx}(1) = 0 $$ Step 6: Combine the differentiated terms. $$ 2xy + x^2\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0 $$ Step 7: Isolate terms containing $\frac{dy}{dx}$ on one side. $$ x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = -2xy - 4 $$ Step 8: Factor out $\frac{dy}{dx}$. $$ \frac{dy}{dx}(x^2 + 2y) = -2xy - 4 $$ Step 9: Solve for $\frac{dy}{dx}$. $$ \frac{dy}{dx} = \frac{-2xy - 4}{x^2 + 2y} $$ $$ \frac{dy}{dx} = -\frac{2(xy + 2)}{x^2 + 2y} $$ The final answer is $\boxed{-\frac{2(xy + 2)}{x^2 + 2y}}$. Send me the next one 📸