Answer: 
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Work Shown:
First calculate
through the use of implicit differentiation.
Don't forget about the chain rule.
![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}\\\\](https://tex.z-dn.net/?f=y%5E5%20%3D%20x%5E7%5C%5C%5C%5C%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5By%5E5%5Cright%5D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5Bx%5E7%5Cright%5D%5C%5C%5C%5C5y%5E4%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%207x%5E6%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B7x%5E6%7D%7B5y%5E4%7D%5C%5C%5C%5C)
Go back to line 3, shown above, and apply the derivative to both sides.
You'll be using the product rule.
![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\\\\](https://tex.z-dn.net/?f=5y%5E4%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%207x%5E6%5C%5C%5C%5C%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B5y%5E4%5Cfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B7x%5E6%5Cright%5D%5C%5C%5C%5C%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B5y%5E4%5Cright%5D%2A%5Cfrac%7Bdy%7D%7Bdx%7D%2B5y%5E4%2A%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%5Cfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%20%3D%2042x%5E5%5C%5C%5C%5C20y%5E3%2A%5Cfrac%7Bdy%7D%7Bdx%7D%2A%5Cfrac%7Bdy%7D%7Bdx%7D%2B5y%5E4%2A%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D42x%5E5%5C%5C%5C%5C20y%5E3%2A%5Cleft%28%5Cfrac%7Bdy%7D%7Bdx%7D%5Cright%29%5E2%2B5y%5E4%2A%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%2042x%5E5%5C%5C%5C%5C)
Use substitution and isolate
to get the following:


