Differentiation Questions

Question Number 163367 by poeter last updated on 06/Jan/22

Answered by som(math1967) last updated on 06/Jan/22

$$\mathrm{1}.\:{equn}\:{of}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{10}\:\left({given}\right) \\$$$$\:\:\frac{{dy}}{{dx}}=\frac{−{x}}{{y}} \\$$$$\left[\frac{{dy}}{{dx}}\right]_{\mathrm{1},\mathrm{3}} =\frac{−\mathrm{1}}{\mathrm{3}} \\$$$$\therefore{slope}\:{of}\:{tanjent}=\frac{−\mathrm{1}}{\mathrm{3}} \\$$$${equation}\:{of}\:{tanjent} \\$$$$\:\:\left({y}−\mathrm{3}\right)=\frac{−\mathrm{1}}{\mathrm{3}}\left({x}−\mathrm{1}\right) \\$$$$\:{x}+\mathrm{3}{y}=\mathrm{10} \\$$

Answered by MikeH last updated on 06/Jan/22

$${f}\left({x},{y}\right)\:=\:\mathrm{4}{x}^{\mathrm{3}} {y}^{\mathrm{2}} \:+\:\mathrm{6}{y}^{\mathrm{3}} \\$$$$\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{12}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:\mathrm{0}\:=\:\mathrm{12}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\$$$$\frac{\partial{f}}{\partial{y}}\:=\:\mathrm{8}{x}^{\mathrm{3}} {y}\:+\:\mathrm{18}{y}^{\mathrm{2}} \\$$