Question

Find y" by implicit differentiation. Simplify as much as possible using substitution when possible. 2x3 5y3 = 7

Answer 1

Answer:

$\frac{d^2y}{dx^2}=\frac{-8x^3-5y^3}{25y^5}$

Step-by-step explanation:

Given, $2x^3+5y^3=7$

   $6x^2+15y^2\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{-6x^2}{15y^2}=\frac{-2x^2}{5y^2}$

Now find the second differentiation w.r.t. $x$.

$\frac{d^2y}{dx^2}=\frac{15y^2(-12x)-(-6x^2)(30y.\frac{dy}{dx})}{225y^4}$

      $=\frac{-180xy^2+180x^2y\frac{dy}{dx}}{225y^4}$

      $=\frac{180xy(x\frac{dy}{dx}-y)}{225y^4}$

      $=\frac{4(x\frac{dy}{dx}-y)}{5y^3}\quad \quad ...(i)$

put value of  $\frac {dy}{dx}$  in equation $(i)$.

 $\frac{d^2y}{dx^2}=\frac{4(x(\frac{-2x^2}{5y^2})-y}{5y^3}$

      $=\frac{\frac{-8x^3}{5y^2}-y}{5y^3}$

$\Rightarrow \frac{d^2y}{dx^2}=\frac{-8x^3-5y^3}{25y^5}$

Hence, $\frac{d^2y}{dx^2}=\frac{-8x^3-5y^3}{25y^5}$.