Question Number 108936 by mohammad17 last updated on 20/Aug/20
Commented by mohammad17 last updated on 20/Aug/20
$${sir}\:{can}\:{you}\:{help}\:{me}\:\left({true}\:{or}\:{false}\right) \\ $$
Answered by Aziztisffola last updated on 20/Aug/20
$$\mathrm{y}=\left(\mathrm{5x}^{\mathrm{2}} \right)^{\mathrm{2}} \Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2}\left(\mathrm{5x}^{\mathrm{2}} \right)×\mathrm{10x}=\mathrm{100x}^{\mathrm{3}} \\ $$$$\mathrm{or}\:\mathrm{y}=\mathrm{25x}^{\mathrm{4}} \:\mathrm{then}\:\mathrm{y}'=\mathrm{4}×\mathrm{25x}^{\mathrm{3}} =\mathrm{100x}^{\mathrm{3}} \\ $$
Commented by mohammad17 last updated on 20/Aug/20
$${very}\:{thank}\:{you}\:{sir} \\ $$