Differentiate the function $$v = \left(\sqrt{x}+\frac 1 {x^{1/3}}\right)^2$$
I tries using the power rule, but it did not work out. Any help would be much appreciated!
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Have you learned of the chain rule? $\frac d {dx} f(g(x)) = \frac{d f} {dg} \cdot \frac{dg}{dx}$ – mathreadler May 04 '17 at 08:00
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1Just expand it. – May 04 '17 at 08:03
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Method 1 : Chain rule says that :
$$\big(f \circ g\big)'=f'(g) \cdot g'$$
Therefore :
$$v' = 2\left(\sqrt{x}+\frac 1 {x^{1/3}}\right)^{2-1} \cdot \left(\sqrt{x}+\frac 1 {x^{1/3}}\right)'=2\left(\sqrt{x}+\frac 1 {x^{1/3}}\right)\cdot \left(\frac{1}{2\sqrt{x}}-\frac {1} {3x^{4/3}}\right)$$$$=\color{blue}{1-\frac{2}{3x^{5/3}}-\frac{1}{3x^{7/6}}}$$
Method 2: $$v = \left(\sqrt{x}+\frac 1 {x^{1/3}}\right)^2=x+\frac1{{x}^{2/3}}+2\sqrt x \frac{1}{x^{2/3}} $$
Therefore ;
$$v' = \left(x+\frac1{{x}^{2/3}}+2x^{-1/6}\right)' =\color{blue}{1-\frac{2}{3x^{5/3}}-\frac{1}{3x^{7/6}}}$$
Jaideep Khare
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