Answer: [tex]\frac{14y}{25x^2}[/tex]
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Work Shown:
First calculate [tex]\frac{dy}{dx}[/tex] through the use of implicit differentiation.
Don't forget about the chain rule.
[tex]y^5 = x^7\\\\\frac{d}{dx}\left[y^5\right] = \frac{d}{dx}\left[x^7\right]\\\\5y^4\frac{dy}{dx} = 7x^6\\\\\frac{dy}{dx} = \frac{7x^6}{5y^4}\\\\[/tex]
Go back to line 3, shown above, and apply the derivative to both sides.
You'll be using the product rule.
[tex]5y^4\frac{dy}{dx} = 7x^6\\\\\frac{d}{dx}\left[5y^4\frac{dy}{dx}\right] = \frac{d}{dx}\left[7x^6\right]\\\\\frac{d}{dx}\left[5y^4\right]*\frac{dy}{dx}+5y^4*\frac{d}{dx}\left[\frac{dy}{dx}\right] = 42x^5\\\\20y^3*\frac{dy}{dx}*\frac{dy}{dx}+5y^4*\frac{d^2y}{dx^2}=42x^5\\\\20y^3*\left(\frac{dy}{dx}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\[/tex]
Use substitution and isolate [tex]\frac{d^2y}{dx^2}[/tex] to get the following:
[tex]20y^3*\left(\frac{dy}{dx}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\20y^3*\left(\frac{7x^6}{5y^4}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\\frac{196x^{12}}{5y^5}+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\\frac{196x^{12}}{5x^7}+5y^4\frac{d^2y}{dx^2} = 42x^5\\\\[/tex]
[tex]\frac{196x^5}{5}+5y^4*\frac{d^2y}{dx^2}=42x^5\\\\5y^4*\frac{d^2y}{dx^2}=42x^5-\frac{196x^5}{5}\\\\5y^4*\frac{d^2y}{dx^2}=\frac{210x^5-196x^5}{5}\\\\5y^4*\frac{d^2y}{dx^2}=\frac{14x^5}{5}\\\\[/tex]
[tex]\frac{d^2y}{dx^2}=\frac{14x^5}{5}*\frac{1}{5y^4}\\\\\frac{d^2y}{dx^2}=\frac{14x^5}{25y^4}\\\\\frac{d^2y}{dx^2}=\frac{14x^5*x^2}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14x^7}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14y^5}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14y}{25x^2}\\\\[/tex]
