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Question Number 94768 by Hamida last updated on 20/May/20

Answered by mathmax by abdo last updated on 21/May/20

$$\mathrm{y}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^4+{6\mathrm{x}}^2+4}{{\mathrm{x}}^2+5}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{({4\mathrm{x}}^3+12\mathrm{x})({\mathrm{x}}^2+5)-2\mathrm{x}({\mathrm{x}}^4+{6\mathrm{x}}^2+4)}{{({\mathrm{x}}^2+5)}^2}$$$$=\frac{{4\mathrm{x}}^5+{20\mathrm{x}}^3+{12\mathrm{x}}^3+60\mathrm{x}-2{\mathrm{x}}^5-{12\mathrm{x}}^3-8\mathrm{x}}{{({\mathrm{x}}^2+5)}^2}$$$$=\frac{{2\mathrm{x}}^5+{20\mathrm{x}}^3+52\mathrm{x}}{{({\mathrm{x}}^2+5)}^2}$$

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