Generating...                               quiz14_n18

Let

$\displaystyle y^3-x^3=4 $

Then answer the following questions.

  1. Find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

    $\displaystyle \frac{dy}{dx} = {{3\,y^2}\over{3\,x^2+4}} $ $\displaystyle \frac{dy}{dx} = -3\,x^2 $ $\displaystyle \frac{dy}{dx} = {{x^2}\over{y^2}} $ $\displaystyle \frac{dy}{dx} = {{y^2}\over{x^2}} $ $\displaystyle \frac{dy}{dx} = {{3\,x^2+4}\over{3\,y^2}} $

  2. Differentiate $\displaystyle \frac{dy}{dx}$.

    $\displaystyle \frac{d^2y}{dx^2} = {{2\,y\,\left(x\,\left({{d}\over{d\,x}}\,y\right)-y\right)}\over{x^ 3}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{2\,x\,\left(x\,\left({{d}\over{d\,x}}\,y\right)-y\right)}\over{y ^3}} $ $\displaystyle \frac{d^2y}{dx^2} = {{6\,y\,\left(3\,x^2\,\left({{d}\over{d\,x}}\... ...({{d }\over{d\,x}}\,y\right)-3\,x\,y\right)}\over{\left(3\,x^2+4\right)^2 }} $ $\displaystyle \frac{d^2y}{dx^2} = -6\,x $ $\displaystyle \frac{d^2y}{dx^2} = -{{2\,\left(3\,x^2\,\left({{d}\over{d\,x}}\,y\right)+4\,\left({{d }\over{d\,x}}\,y\right)-3\,x\,y\right)}\over{3\,y^3}} $

  3. Find $\displaystyle \frac{d^2y}{dx^2}$ by substituting $\displaystyle \frac{d}{dx}y$

    $\displaystyle \frac{d^2y}{dx^2} = -6\,x $ $\displaystyle \frac{d^2y}{dx^2} = {{18\,x\,y^3-18\,x^4-48\,x^2-32}\over{9\,y^5}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{18\,x\,y^3-18\,x^4-48\,x^2-32}\over{\left(9\,x^4+24\,x^2+16 \right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{2\,y^3-2\,x^3}\over{x^3\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{2\,x\,y^3-2\,x^4}\over{y^5}} $

  4. Simplify $\displaystyle \frac{d^2y}{dx^2}$ if possible.

    $\displaystyle \frac{d^2y}{dx^2} = {{48\,x^2-72\,x+32}\over{\left(9\,x^4+24\,x^2+16\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{8}\over{x^3\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{8\,x}\over{y^5}} $ $\displaystyle \frac{d^2y}{dx^2} = -6\,x $ $\displaystyle \frac{d^2y}{dx^2} = -{{48\,x^2-72\,x+32}\over{9\,y^5}} $

  5. Find $\displaystyle \frac{d^2y}{dx^2}$ if $\displaystyle y^3-x^3=5$ .

    $\displaystyle \frac{d^2y}{dx^2} = {{48\,x^2-90\,x+32}\over{\left(9\,x^4+24\,x^2+16\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -6\,x $ $\displaystyle \frac{d^2y}{dx^2} = {{10\,x}\over{y^5}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{10}\over{x^3\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{48\,x^2-90\,x+32}\over{9\,y^5}} $


Department of Mathematics
Last modified: 2026-02-06