Generating...                               quiz14_n6

Let

$\displaystyle y^2-x^2=1 $

Then answer the following questions.

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

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

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

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

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

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

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

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

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

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


Department of Mathematics
Last modified: 2025-06-19