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Check the form of the limit. As x infinity, both the numerator (x² - x) and the denominator (x³ + 2x) approach infinity. This is an indeterminate form (infinity)/(infinity), so L'Hôpital's

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3 parts
a

−x+1x21+3x3-\frac{x+1}{x^2 \sqrt[3]{1+3x}}

b

−32xcsc⁡(34x2)cot⁡(34x2)-\frac{3}{2}x \csc\left(\frac{3}{4}x^2\right)\cot\left(\frac{3}{4}x^2\right)

c

−2(xy+2)x2+2y-\frac{2(xy + 2)}{x^2 + 2y}

Step-by-step solution

19 steps

  1. To evaluate lim⁡x→∞x2−xx3+2x\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→∞x \to \infty, both the numerator (x2−x)(x^2 - x) and the denominator (x3+2x)(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.

    ddx(x2−x)=2x−1\frac{d}{dx}(x^2 - x) = 2x - 1 ddx(x3+2x)=3x2+2\frac{d}{dx}(x^3 + 2x) = 3x^2 + 2

    Step 3: Apply L'Hôpital's rule.

    lim⁡x→∞x2−xx3+2x=lim⁡x→∞2x−13x2+2\lim_{x \to \infty} \frac{x^2 - x}{x^3 + 2x} = \lim_{x \to \infty} \frac{2x - 1}{3x^2 + 2}

    Again, as x→∞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.

    ddx(2x−1)=2\frac{d}{dx}(2x - 1) = 2 ddx(3x2+2)=6x\frac{d}{dx}(3x^2 + 2) = 6x

    Step 5: Apply L'Hôpital's rule a second time.

    lim⁡x→∞2x−13x2+2=lim⁡x→∞26x\lim_{x \to \infty} \frac{2x - 1}{3x^2 + 2} = \lim_{x \to \infty} \frac{2}{6x}

    Step 6: Evaluate the limit.
    As x→∞x \to \infty, 6x→∞6x \to \infty, so 26x→0\frac{2}{6x} \to 0.

    lim⁡x→∞26x=0\lim_{x \to \infty} \frac{2}{6x} = 0

    The final answer is 0\boxed{0}.

  2. Differentiate the following functions:

Part a· 6 steps

y=(1+3x)23xy = \frac{\sqrt[3]{(1+3x)^2}}{x}

  1. Rewrite the function using exponents.

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

  2. Apply the quotient rule, ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.

    Let $u = (1+3x)^{2/3}$ and $v = x$.

  3. Find u′u' using the chain rule.

    $$
u' = \frac{2}{3}(1+3x)^{(2/3)-1} \cdot \frac{d}{dx}(1+3x)
$$
$$
u' = \frac{2}{3}(1+3x)^{-1/3} \cdot 3
$$
$$
u' = 2(1+3x)^{-1/3}
$$

  4. Find v′v'.

    $$
v' = \frac{d}{dx}(x) = 1
$$

  5. Substitute $u, v, u', v'$ into the quotient rule formula.

    $$
\frac{dy}{dx} = \frac{2(1+3x)^{-1/3} \cdot x - (1+3x)^{2/3} \cdot 1}{x^2}
$$

  6. Simplify the expression.

    Factor out (1+3x)−1/3(1+3x)^{-1/3} from the numerator. dydx=(1+3x)−1/3[2x−(1+3x)2/3(1+3x)1/3]x2 \frac{dy}{dx} = \frac{(1+3x)^{-1/3} [2x - (1+3x)^{2/3}(1+3x)^{1/3}]}{x^2} dydx=(1+3x)−1/3[2x−(1+3x)]x2 \frac{dy}{dx} = \frac{(1+3x)^{-1/3} [2x - (1+3x)]}{x^2} dydx=(1+3x)−1/3[2x−1−3x]x2 \frac{dy}{dx} = \frac{(1+3x)^{-1/3} [2x - 1 - 3x]}{x^2} dydx=(1+3x)−1/3[−x−1]x2 \frac{dy}{dx} = \frac{(1+3x)^{-1/3} [-x - 1]}{x^2} dydx=−(x+1)x2(1+3x)1/3 \frac{dy}{dx} = \frac{-(x+1)}{x^2 (1+3x)^{1/3}} The final answer is −x+1x21+3x3\boxed{-\frac{x+1}{x^2 \sqrt[3]{1+3x}}}.

Part b

−32xcsc⁡(34x2)cot⁡(34x2)-\frac{3}{2}x \csc\left(\frac{3}{4}x^2\right)\cot\left(\frac{3}{4}x^2\right)

y=csc⁡(34x2)y = \csc\left(\frac{3}{4}x^2\right)

  1. Apply the chain rule.

    The derivative of csc⁡(u)\csc(u) is −csc⁡(u)cot⁡(u)⋅u′-\csc(u)\cot(u) \cdot u'.
    Let u=34x2u = \frac{3}{4}x^2.

  2. Find u′u'.

    $$
u' = \frac{d}{dx}\left(\frac{3}{4}x^2\right) = \frac{3}{4} \cdot 2x = \frac{3}{2}x
$$

  3. Substitute uu and u′u' into the chain rule formula.

    $$
\frac{dy}{dx} = -\csc\left(\frac{3}{4}x^2\right)\cot\left(\frac{3}{4}x^2\right) \cdot \left(\frac{3}{2}x\right)
$$

  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 $\boxed{-\frac{3}{2}x \csc\left(\frac{3}{4}x^2\right)\cot\left(\frac{3}{4}x^2\right)}$.

Part c

−2(xy+2)x2+2y-\frac{2(xy + 2)}{x^2 + 2y}

x2y+y2+4x=1x^2y + y^2 + 4x = 1

  1. Differentiate both sides of the equation with respect to xx.

    Remember to use implicit differentiation for terms involving yy.

  2. Differentiate x2yx^2y using the product rule.

    $$
\frac{d}{dx}(x^2y) = \frac{d}{dx}(x^2) \cdot y + x^2 \cdot \frac{d}{dx}(y) = 2xy + x^2\frac{dy}{dx}
$$

  3. Differentiate y2y^2 using the chain rule.

    $$
\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}
$$

  4. Differentiate $4x$.

    $$
\frac{d}{dx}(4x) = 4
$$

  5. Differentiate the constant 1.

    $$
\frac{d}{dx}(1) = 0
$$

  6. Combine the differentiated terms.

    $$
2xy + x^2\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0
$$

  7. Isolate terms containing dydx\frac{dy}{dx} on one side.

    $$
x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = -2xy - 4
$$

  8. Factor out dydx\frac{dy}{dx}.

    $$
\frac{dy}{dx}(x^2 + 2y) = -2xy - 4
$$

  9. Solve for dydx\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 $\boxed{-\frac{2(xy + 2)}{x^2 + 2y}}$.
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