Solve using basic differentiation rule
š(š„) =3ā2š„/3+2š„
ddx(3ā2x3+2x)=gā(x)\frac{d}{dx}(3-\frac{2x}{3}+2x)=gā(x)dxdā(3ā32xā+2x)=gā(x)
gā(x)=ā23+2=43gā(x)=-\frac{2}{3}+2=\frac{4}{3}gā(x)=ā32ā+2=34ā
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