Question 185259
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f(x) = y = \frac{2}{sqrt{x}} + \frac{x}{sqrt{3}} + \frac{6} {\sqrt[3]{x^2}}]


Since we know that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \phi(x) = \alpha(x) + \beta(x) \ \ \Rightarrow\ \ \phi'(x) = \alpha'(x) + \beta'(x)]


We can take the derivitive of each term of your function and just add them.  First thing to do is put your functions in terms of fractional exponents using the rules:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt[m]{a^n} = a^{\frac{n}{m}}]


and 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{a^n} = a^{-n}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x) = \frac{2}{sqrt{x}} \ \ \Rightarrow\ \ g(x) = 2x^{-\frac{1}{2}}]


Now take the derivitive using the Power Rule: *[tex \Large \frac{d(ax^n)}{dx} = anx^{n-1}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x) = 2x^{-\frac{1}{2}}\ \ \Rightarrow\ \ g'(x) = 2\left(-\frac{1}{2}\right)x^{-\frac{1}{2}-1}\  = -x^{-\frac{3}{2}}\ =\frac{-1}{sqrt{x^3}} ]


Look at the next term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(x) = \frac{x}{sqrt{3}} \ \ \Rightarrow\ \ h(x) = \frac{1}{sqrt{3}} x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  h'(x) = \frac{1}{sqrt{3}}\ \cdot\ 1\ x^{1 - 1}\ = \frac{1}{sqrt{3}}x^0\ =  \frac{1}{sqrt{3}} ]


And then the last term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k(x) = \frac{6} {\sqrt[3]{x^2}} = 6x^{-\frac{2}{3}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k'(x) = 6\left(-\frac{2}{3}\right)x^{\left(-\frac{2}{3}-1\right)}\  = -4x^{-\frac{5}{3}} = -\frac{4}{\sqrt[3]{x^5}} = -\frac{4}{x\sqrt[3]{x^2}} ]


Then put it all back together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x) = g'(x) + h'(x) + k'(x) = \frac{-1}{sqrt{x^3}} + \frac{1}{sqrt{3}} -\frac{4}{x\sqrt[3]{x^2}} ]


However, I would rationalize all the denominators:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x) = -\frac{sqrt{x^3}}{x^3} + \frac{sqrt{3}}{3} - \frac{4 \sqrt[3]{x}}{x^2}]





John
*[tex \LARGE e^{i\pi} + 1 = 0]
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