Question Number 25528 by Mahesh Andiboina last updated on 11/Dec/17
Commented by Mahesh Andiboina last updated on 11/Dec/17
$$\mathrm{plz}\:\mathrm{ans}\:\mathrm{both} \\ $$
Answered by sushmitak last updated on 11/Dec/17
$${y}=\frac{\mathrm{1}}{\:\sqrt{{x}}}={x}^{−\mathrm{1}/\mathrm{2}} \\ $$$$\frac{{d}}{{dx}}{y}=−\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} =−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}}} \\ $$
Commented by prakash jain last updated on 11/Dec/17
$$\frac{{d}}{{dx}}{x}^{{n}} ={nx}^{{n}−\mathrm{1}} \\ $$$$\frac{{d}}{{dx}}{x}^{−\mathrm{1}/\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\mathrm{3}/\mathrm{2}} \\ $$
Commented by ibraheem160 last updated on 11/Dec/17
$${i}\:{guessed}\:{the}\:{answer}\:{should}\:{be}\:\:−\frac{\mathrm{3}}{\mathrm{2}\sqrt{{x}}} \\ $$
Answered by sushmitak last updated on 11/Dec/17
$$\frac{{dy}}{{dx}}=\frac{{dy}/{dt}}{{dx}/{dt}}=\frac{\frac{{d}}{{dt}}\left({t}^{\mathrm{3}} −{t}\right)}{\frac{{d}}{{dt}}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{6}{t}} \\ $$