Find y′ of y=13√x+√x
y′=ddx(13√x+√x)y′=(3√x+√x)ddx(1)−(1)ddx(3√x+√x)(3√x+√x)2y′=(x+√x)13ddx(1)−(1)ddx(x+√x)13(3√x+√x)2y′=(x+√x)13(0)−(1)(13)(x+√x)−23ddx(x+√x)(x+√x)23y′=−13(x+√x)−23(1+12√x)(x+√x)23y′=−(2√x+1)3(2√x)(x+√x)23(x+√x)23y′=−(2√x+1)6√x(x+√x)43
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