Question

Find d²y/dx² in terms of x and y.y^5 = x^7d²y/dx² =????​

Answer 1

Answer:   $\frac{14y}{25x^2}$

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Work Shown:

First calculate $\frac{dy}{dx}$ through the use of implicit differentiation.

Don't forget about the chain rule.

$y^5 = x^7\\\\\frac{d}{dx}\left[y^5\right] = \frac{d}{dx}\left[x^7\right]\\\\5y^4\frac{dy}{dx} = 7x^6\\\\\frac{dy}{dx} = \frac{7x^6}{5y^4}$

Go back to line 3, shown above, and apply the derivative to both sides.

You'll be using the product rule.

$5y^4\frac{dy}{dx} = 7x^6\\\\\frac{d}{dx}\left[5y^4\frac{dy}{dx}\right] = \frac{d}{dx}\left[7x^6\right]\\\\\frac{d}{dx}\left[5y^4\right]*\frac{dy}{dx}+5y^4*\frac{d}{dx}\left[\frac{dy}{dx}\right] = 42x^5\\\\20y^3*\frac{dy}{dx}*\frac{dy}{dx}+5y^4*\frac{d^2y}{dx^2}=42x^5\\\\20y^3*\left(\frac{dy}{dx}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5$

Use substitution and isolate $\frac{d^2y}{dx^2}$ to get the following:

$20y^3*\left(\frac{dy}{dx}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\20y^3*\left(\frac{7x^6}{5y^4}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\\frac{196x^{12}}{5y^5}+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\\frac{196x^{12}}{5x^7}+5y^4\frac{d^2y}{dx^2} = 42x^5$

$\frac{196x^5}{5}+5y^4*\frac{d^2y}{dx^2}=42x^5\\\\5y^4*\frac{d^2y}{dx^2}=42x^5-\frac{196x^5}{5}\\\\5y^4*\frac{d^2y}{dx^2}=\frac{210x^5-196x^5}{5}\\\\5y^4*\frac{d^2y}{dx^2}=\frac{14x^5}{5}$

$\frac{d^2y}{dx^2}=\frac{14x^5}{5}*\frac{1}{5y^4}\\\\\frac{d^2y}{dx^2}=\frac{14x^5}{25y^4}\\\\\frac{d^2y}{dx^2}=\frac{14x^5*x^2}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14x^7}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14y^5}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14y}{25x^2}$