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

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

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

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

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

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

    $\displaystyle \frac{d^2y}{dx^2} = 2 $ $\displaystyle \frac{d^2y}{dx^2} = {{1}\over{x^2\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{4\,x-5}\over{\left(4\,x^2-4\,x+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = {{4\,x-5}\over{4\,y^3}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{1}\over{y^3}} $

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

    $\displaystyle \frac{d^2y}{dx^2} = 2 $ $\displaystyle \frac{d^2y}{dx^2} = {{4\,x-9}\over{4\,y^3}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{4\,x-9}\over{\left(4\,x^2-4\,x+1\right)\,y}} $ $\displaystyle \frac{d^2y}{dx^2} = -{{2}\over{y^3}} $ $\displaystyle \frac{d^2y}{dx^2} = {{2}\over{x^2\,y}} $


Department of Mathematics
Last modified: 2026-08-21