# Simplifying the second derivative I need to simplify the below with respect to y′′. I'm given:y''=-(3x^2y^3-3x^3y^2(-(x^3)/(y^3)))/(y^6} The final result should look like: -(3x^2(y^4+x^4))/(y^7)

Simplifying the second derivative
I need to simplify the below with respect to ${y}^{″}$. I'm given:
${y}^{″}=-\frac{3{x}^{2}{y}^{3}-3{x}^{3}{y}^{2}\left(-\frac{{x}^{3}}{{y}^{3}}\right)}{{y}^{6}}$
The final result should look like:
$-\frac{3{x}^{2}\left({y}^{4}+{x}^{4}\right)}{{y}^{7}}$
Here are my steps:
1.Combine term: $-3{x}^{3}{y}^{2}\left(-\frac{{x}^{3}}{{y}^{3}}\right)$
$-3{x}^{3}{y}^{2}\left(-\frac{{x}^{3}}{{y}^{3}}\right)=\frac{3{x}^{6}{y}^{2}}{{y}^{3}}$
Our expression is now:
${y}^{″}=-\frac{3{x}^{2}{y}^{3}+\frac{3{x}^{6}{y}^{2}}{{y}^{3}}}{{y}^{6}}$
2.Multiply the final term in the numerator, $\frac{3{x}^{6}{y}^{2}}{{y}^{3}}$, by the it's reciprocal in the denominator, $\frac{{y}^{6}}{1}$, to eliminate the complex fraction.
$\frac{3{x}^{6}{y}^{2}}{{y}^{3}}\ast \frac{{y}^{6}}{1}=\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$
Our expression is now:
${y}^{″}=-\frac{3{x}^{2}{y}^{3}}{{y}^{6}}+\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$
3.Multiply term $\frac{3{x}^{6}{y}^{8}}{{y}^{3}}$ by $\frac{{y}^{3}}{{y}^{3}}$ to make common denominators.
$\frac{3{x}^{6}{y}^{8}}{{y}^{3}}\ast \frac{{y}^{3}}{{y}^{3}}=\frac{3{x}^{6}{y}^{11}}{{y}^{6}}$
Our separate terms can now be added, and our expression will look like:
${y}^{″}=-\frac{3{x}^{2}{y}^{3}+3{x}^{6}{y}^{11}}{{y}^{6}}$
4.Now factor the numerator, our expression will become:
${y}^{″}=-\frac{3{x}^{2}{y}^{3}\left(1+{x}^{4}{y}^{8}\right)}{{y}^{6}}$
Have I made a mistake? I'm not sure where to go from here. I need to achieve this expression as stated at the top my question:
${y}^{″}=-\frac{3{x}^{2}\left({y}^{4}+{x}^{4}\right)}{{y}^{7}}$
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Kitamiliseakekw
${y}^{″}=-\frac{3{x}^{2}{y}^{3}-3{x}^{3}{y}^{2}\left(-\frac{{x}^{3}}{{y}^{3}}\right)}{{y}^{6}}=-\frac{3{x}^{2}{y}^{3}+3{x}^{3}\left(\frac{{x}^{3}}{y}\right)}{{y}^{6}}={y}^{″}=-\frac{\left(\frac{3{x}^{2}{y}^{4}+3{x}^{6}}{y}\right)}{{y}^{6}}=-\frac{3{x}^{2}{y}^{4}+3{x}^{6}}{{y}^{7}}=-\frac{3{x}^{2}\left({y}^{4}+{x}^{4}\right)}{{y}^{7}}$

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