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

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

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

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

    $\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}} $ $\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\,y^3-2\,x^3}\over{x^3\,y}} $

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

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

  5. Find $\displaystyle \frac{d^2y}{dx^2}$ if $\displaystyle y^3-x^3=4$ .

    $\displaystyle \frac{d^2y}{dx^2} = -6\,x $ $\displaystyle \frac{d^2y}{dx^2} = -{{8}\over{x^3\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{4\,x^2-8\,x+2}\over{\left(x^4+2\,x^2+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{8\,x}\over{y^5}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{4\,x^2-8\,x+2}\over{y^5}} $


Department of Mathematics
Last modified: 2025-08-14