Question Number 143598 by mohammad17 last updated on 16/Jun/21
Commented by mohammad17 last updated on 16/Jun/21
$${help}\:{me}\:{sir} \\ $$
Commented by mohammad17 last updated on 16/Jun/21
$$??????? \\ $$
Answered by qaz last updated on 16/Jun/21
$$\mathrm{y}=\mathrm{3z}^{\mathrm{3}} +\mathrm{2}......\Rightarrow\mathrm{z}^{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{z}^{\mathrm{2}} +\mathrm{4}.......\Rightarrow\mathrm{z}^{\mathrm{6}} =\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{9}\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{y}−\mathrm{2}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{27}\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{27}\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{y}−\mathrm{2}\right)} \\ $$
Answered by bobhans last updated on 16/Jun/21
$$\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dz}}×\frac{{dz}}{{dx}} \\ $$
Answered by mr W last updated on 16/Jun/21
$$\frac{{dy}}{{dz}}=\mathrm{9}{z}^{\mathrm{2}} \\ $$$$\frac{{dx}}{{dz}}=\mathrm{2}{z} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{9}{z}}{\mathrm{2}}=\frac{\mathrm{3}\left({y}−\mathrm{2}\right)}{\mathrm{2}\left({x}−\mathrm{4}\right)} \\ $$