Answer:
[tex]\frac{d^2y}{dx^2}=\frac{-8x^3-5y^3}{25y^5}[/tex]
Step-by-step explanation:
Given, [tex]2x^3+5y^3=7[/tex]
[tex]6x^2+15y^2\frac{dy}{dx}=0[/tex]
[tex]\Rightarrow \frac{dy}{dx}=\frac{-6x^2}{15y^2}=\frac{-2x^2}{5y^2}[/tex]
Now find the second differentiation w.r.t. [tex]x[/tex].
[tex]\frac{d^2y}{dx^2}=\frac{15y^2(-12x)-(-6x^2)(30y.\frac{dy}{dx})}{225y^4}[/tex]
[tex]=\frac{-180xy^2+180x^2y\frac{dy}{dx}}{225y^4}[/tex]
[tex]=\frac{180xy(x\frac{dy}{dx}-y)}{225y^4}[/tex]
[tex]=\frac{4(x\frac{dy}{dx}-y)}{5y^3}\quad \quad ...(i)[/tex]
put value of [tex]\frac {dy}{dx}[/tex] in equation [tex](i)[/tex].
[tex]\frac{d^2y}{dx^2}=\frac{4(x(\frac{-2x^2}{5y^2})-y}{5y^3}[/tex]
[tex]=\frac{\frac{-8x^3}{5y^2}-y}{5y^3}[/tex]
[tex]\Rightarrow \frac{d^2y}{dx^2}=\frac{-8x^3-5y^3}{25y^5}[/tex]
Hence, [tex]\frac{d^2y}{dx^2}=\frac{-8x^3-5y^3}{25y^5}[/tex].
