Question #340775

Solve using basic differentiation rule


y = 2/(x ^ (1/2)) + 6/(x ^ (1/3)) - 2/(x ^ (3/2)) + 4/(x ^ (3/4))

1
Expert's answer
2022-05-17T09:57:57-0400

y=2x1/2+6x1/32x3/2+4x3/4=2x1/2+6x1/32x3/2+4x3/4y=\frac{2}{x^{1/2}}+\frac{6}{x^{1/3}}-\frac{2}{x^{3/2}}+\frac{4}{x^{3/4}}=2x^{-1/2}+6x^{-1/3}-2x^{-3/2}+4x^{-3/4}

y=x3/22x4/3+3x5/23x7/4=1x3/22x4/3+3x5/23x7/4y'=-x^{-3/2}-2x^{-4/3}+3x^{-5/2}-3x^{-7/4}=-\frac{1}{x^{3/2}}-\frac{2}{x^{4/3}}+\frac{3}{x^{5/2}}-\frac{3}{x^{7/4}}







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