Generating...                               quiz14_n3

Let

$\displaystyle y^3+x^3=3 $

Then answer the following questions.

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

    $\displaystyle \frac{dy}{dx} = -{{y^2}\over{x^2-1}} $ $\displaystyle \frac{dy}{dx} = -{{x^2-1}\over{y^2}} $ $\displaystyle \frac{dy}{dx} = -{{y^2}\over{x^2}} $ $\displaystyle \frac{dy}{dx} = -{{x^2}\over{y^2}} $ $\displaystyle \frac{dy}{dx} = 3\,x^2 $

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

    $\displaystyle \frac{d^2y}{dx^2} = -{{2\,y\,\left(x^2\,\left({{d}\over{d\,x}}\,y... ...d}\over{d\,x }}\,y-x\,y\right)}\over{\left(x-1\right)^2\,\left(x+1\right)^2}} $ $\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} = {{2\,\left(x^2\,\left({{d}\over{d\,x}}\,y\right)-{{d}\over{d\,x}}\, y-x\,y\right)}\over{y^3}} $ $\displaystyle \frac{d^2y}{dx^2} = 6\,x $ $\displaystyle \frac{d^2y}{dx^2} = -{{2\,y\,\left(x\,\left({{d}\over{d\,x}}\,y\right)-y\right)}\over{x ^3}} $

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

    $\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^4-4\,x^2+2}\over{\left(x^4-2\,x^2+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = 6\,x $ $\displaystyle \frac{d^2y}{dx^2} = -{{2\,x\,y^3+2\,x^4-4\,x^2+2}\over{y^5}} $ $\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} = -{{6\,x}\over{y^5}} $ $\displaystyle \frac{d^2y}{dx^2} = 6\,x $ $\displaystyle \frac{d^2y}{dx^2} = {{6}\over{x^3\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{4\,x^2-6\,x-2}\over{\left(x^4-2\,x^2+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{4\,x^2-6\,x-2}\over{y^5}} $

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

    $\displaystyle \frac{d^2y}{dx^2} = -{{4\,x^2-4\,x-2}\over{\left(x^4-2\,x^2+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{4}\over{x^3\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{4\,x^2-4\,x-2}\over{y^5}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{4\,x}\over{y^5}} $ $\displaystyle \frac{d^2y}{dx^2} = 6\,x $


Department of Mathematics
Last modified: 2026-03-24