Generating...                               quiz14_n29

Let

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

Then answer the following questions.

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

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

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

    $\displaystyle \frac{d^2y}{dx^2} = -{{3\,\left(4\,x^3\,\left({{d}\over{d\,x}}\,y\right)+3\,\left({{d }\over{d\,x}}\,y\right)-4\,x^2\,y\right)}\over{4\,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\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = {{12\,y^2\,\left(4\,x^3\,\left({{d}\over{d\,x... ... {{d}\over{d\,x}}\,y\right)-4\,x^2\,y\right)}\over{\left(4\,x^3+3 \right)^2}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{3\,x^2\,\left(x\,\left({{d}\over{d\,x}}\,y\right)-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} = {{48\,x^2\,y^4-48\,x^6-72\,x^3-27}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{48\,x^2\,y^4-48\,x^6-72\,x^3-27}\over{\left(16\,x^6+24\,x^3+9 \right)\,y}} $

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

    $\displaystyle \frac{d^2y}{dx^2} = {{72\,x^3-144\,x^2+27}\over{\left(16\,x^6+24\,x^3+9\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{72\,x^3-144\,x^2+27}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{9}\over{x^4\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -12\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = {{9\,x^2}\over{y^7}} $

  5. Find $\displaystyle \frac{d^2y}{dx^2}$ if $\displaystyle y^4-x^4=1$ .

    $\displaystyle \frac{d^2y}{dx^2} = -12\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = {{3\,x^2}\over{y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{72\,x^3-48\,x^2+27}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = {{72\,x^3-48\,x^2+27}\over{\left(16\,x^6+24\,x^3+9\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{3}\over{x^4\,y}} $


Department of Mathematics
Last modified: 2025-10-30