Let

$\displaystyle y^4-x^4=1 $

Then answer the following questions.

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

    $\displaystyle \frac{dy}{dx} = {{4\,y^3}\over{4\,x^3+1}} $ $\displaystyle \frac{dy}{dx} = {{x^3}\over{y^3}} $ $\displaystyle \frac{dy}{dx} = {{y^3}\over{x^3}} $ $\displaystyle \frac{dy}{dx} = -4\,x^3 $ $\displaystyle \frac{dy}{dx} = {{4\,x^3+1}\over{4\,y^3}} $

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

    $\displaystyle \frac{d^2y}{dx^2} = -12\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = -{{3\,\left(4\,x^3\,\left({{d}\over{d\,x}}\,y\right)+{{d}\over{d\,x }}\,y-4\,x^2\,y\right)}\over{4\,y^4}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{3\,x^2\,\left(x\,\left({{d}\over{d\,x}}\,y\right)-y\right) }\over{y^4}} $ $\displaystyle \frac{d^2y}{dx^2} = {{3\,y^2\,\left(x\,\left({{d}\over{d\,x}}\,y\right)-y\right)}\over{ x^4}} $ $\displaystyle \frac{d^2y}{dx^2} = {{12\,y^2\,\left(4\,x^3\,\left({{d}\over{d\,x... ...\right)+{{d }\over{d\,x}}\,y-4\,x^2\,y\right)}\over{\left(4\,x^3+1\right)^2}} $

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

    $\displaystyle \frac{d^2y}{dx^2} = {{48\,x^2\,y^4-48\,x^6-24\,x^3-3}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{3\,y^4-3\,x^4}\over{x^4\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{3\,x^2\,y^4-3\,x^6}\over{y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{48\,x^2\,y^4-48\,x^6-24\,x^3-3}\over{\left(16\,x^6+8\,x^3+1 \right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -12\,x^2 $

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

    $\displaystyle \frac{d^2y}{dx^2} = {{24\,x^3-48\,x^2+3}\over{\left(16\,x^6+8\,x^3+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{3\,x^2}\over{y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{24\,x^3-48\,x^2+3}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{3}\over{x^4\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -12\,x^2 $

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

    $\displaystyle \frac{d^2y}{dx^2} = -{{24\,x^3-144\,x^2+3}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -12\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = -{{9}\over{x^4\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{24\,x^3-144\,x^2+3}\over{\left(16\,x^6+8\,x^3+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{9\,x^2}\over{y^7}} $


Department of Mathematics
Last modified: 2026-01-25