Generating...                               quiz14_n22

Let

$\displaystyle y^4+x^4=5 $

Then answer the following questions.

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

    $\displaystyle \frac{dy}{dx} = -{{y^3}\over{x^3}} $ $\displaystyle \frac{dy}{dx} = -{{x^3}\over{y^3}} $ $\displaystyle \frac{dy}{dx} = -{{4\,x^3-5}\over{4\,y^3}} $ $\displaystyle \frac{dy}{dx} = -{{4\,y^3}\over{4\,x^3-5}} $ $\displaystyle \frac{dy}{dx} = 4\,x^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)-5\,\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\,y^2\,\left(4\,x^3\,\left({{d}\over{d\,... ...({{d}\over{d\,x}}\,y\right)-4\,x^2\,y\right)}\over{\left(4\,x^3 -5\right)^2}} $ $\displaystyle \frac{d^2y}{dx^2} = 12\,x^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} = 12\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = -{{48\,x^2\,y^4+48\,x^6-120\,x^3+75}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = {{48\,x^2\,y^4+48\,x^6-120\,x^3+75}\over{\left(16\,x^6-40\,x^3+25 \right)\,y}} $ $\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}} $

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

    $\displaystyle \frac{d^2y}{dx^2} = -{{15\,x^2}\over{y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = {{15}\over{x^4\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{120\,x^3-240\,x^2-75}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = 12\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = -{{120\,x^3-240\,x^2-75}\over{\left(16\,x^6-40\,x^3+25\right)\,y}} $

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

    $\displaystyle \frac{d^2y}{dx^2} = -{{12\,x^2}\over{y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = 12\,x^2 $ $\displaystyle \frac{d^2y}{dx^2} = {{120\,x^3-192\,x^2-75}\over{16\,y^7}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{120\,x^3-192\,x^2-75}\over{\left(16\,x^6-40\,x^3+25\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{12}\over{x^4\,y}} $


Department of Mathematics
Last modified: 2026-03-24