Differentiate g(x)=(5−x)2(2x−1)5.
By using Product Rule and Chain Rule,
g′(x)=(5−x)2⋅ddx(2x−1)5+(2x−1)5⋅ddx(5−x)2g′(x)=(5−x)2⋅5(2x−1)5−1⋅ddx(2x−1)+(2x−1)5⋅2(5−x)2−1⋅ddx(5−x)g′(x)=(5−x)2⋅5(2x−1)4(2)+(2x−1)5⋅2(5−x)(−1)g′(x)=10(5−x)2(2x−1)4−2(5−x)(2x−1)5g′(x)=2(5−x)(2x−1)4[5(5−x)−(2x−1)]g′(x)=2(5−x)(2x−1)4[25−5x−2x+1]g′(x)=2(5−x)(2x−1)4(26−7x)g′(x)=2(5−x)(26−7x)(2x−1)4
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