A) \[\frac{2}{27}{{x}^{2}}{{\left( 2{{x}^{3}}+15 \right)}^{3}}\]
B) \[\frac{2}{27}{{x}^{2}}{{\left( 2{{x}^{3}}+15 \right)}^{3}}\]
C) \[\frac{2}{27}x{{\left( 2{{x}^{3}}+5 \right)}^{3}}\]
D) \[\frac{2}{27}{{\left( 2{{x}^{3}}+15 \right)}^{3}}\]
Correct Answer: A
Solution :
We have, \[y=\frac{1}{4}{{u}^{4}}\Rightarrow \frac{dy}{du}=\frac{1}{4}.\,4{{u}^{3}}={{u}^{3}}\] And \[u=\frac{2}{3}{{x}^{3}}+5\] \[\Rightarrow \,\frac{du}{dx}=\frac{2}{3}\,.\,3{{x}^{2}}=2{{x}^{2}}\] \[\therefore \,\,\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}={{u}^{3}}.2{{x}^{2}}={{\left( \frac{2}{3}{{x}^{3}}+5 \right)}^{3}}\left( 2{{x}^{2}} \right)\] \[=\frac{2}{27}{{x}^{3}}{{\left( 2{{x}^{3}}+15 \right)}^{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
