SOLUTION: Differentiate (find the derivative) of the function 2/(√x) + x/(√3) + 6(3√x <-- cube root over x), the answer should be 1/(√3) - 1/√(x^3) - 2/(3√
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Question 185259: Differentiate (find the derivative) of the function 2/(√x) + x/(√3) + 6(3√x <-- cube root over x), the answer should be 1/(√3) - 1/√(x^3) - 2/(3√x^2) <-- cube root over x^2). Please show your steps.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
 = y = \frac{2}{sqrt{x}} + \frac{x}{sqrt{3}} + \frac{6} {\sqrt[3]{x^2}})
Since we know that:
 = \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:
and
 = \frac{2}{sqrt{x}} \ \ \Rightarrow\ \ g(x) = 2x^{-\frac{1}{2}})
Now take the derivitive using the Power Rule: }{dx} = anx^{n-1})
 = 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:
 = \frac{x}{sqrt{3}} \ \ \Rightarrow\ \ h(x) = \frac{1}{sqrt{3}} 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:
 = \frac{6} {\sqrt[3]{x^2}} = 6x^{-\frac{2}{3}})
 = 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:
 = 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:
 = -\frac{sqrt{x^3}}{x^3} + \frac{sqrt{3}}{3} - \frac{4 \sqrt[3]{x}}{x^2})
John
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