Generating...                               quiz14_n19

Let

$\displaystyle y^4+x^4=4 $

Then answer the following questions.

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

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

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

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

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

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

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

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

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

    $\displaystyle \frac{d^2y}{dx^2} = -{{6\,x^3-15\,x^2-3}\over{\left(x^6-2\,x^3+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{15\,x^2}\over{y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = 12\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = {{15}\over{x^4\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{6\,x^3-15\,x^2-3}\over{y^7}} $


Department of Mathematics
Last modified: 2026-07-16