Answer to Question #98912 in Calculus for Ojugbele Daniel

Question

Differentiate cube root of x plus cube root of x

Answer 1

Differentiation

We need to differentiate the Function $\sqrt[3]{x + \sqrt[3]{x}}$

Solution:

We know,

$$Derivative \space of \space x^n = n \times x^{n-1}$$

Chain rule,

$Derivative of f(g(x)) = f ' (g(x)) \times g'(x) \times Derivative \space of \space (x)$

Let the given function as

$$y = \sqrt[3]{x + \sqrt[3]{x}} = ( x + {x}^ {\frac {1}{3}})^{\frac {1}{3}}$$

Now, Differentiate with respect to x,

$$\frac {dy}{dx} = \frac {d}{dx} ( ( x + {x}^ {\frac {1}{3}})^{\frac {1}{3}})$$

$$= \frac {1}{3} \times ( x+ x^{\frac {1}{3}})^{(\frac {1}{3} - 1)} \times \frac {d}{dx} (x + x^{\frac{1}{3}})$$

$$= \frac {1}{3} \times ( x+ x^{\frac {1}{3}})^{(\frac {-2}{3} )} \times ( 1 + \frac {1}{3} \times x ^{\frac {-2}{3}})$$

$$= \frac {1}{9} \times ( x+ x^{\frac {1}{3}})^{(\frac {-2}{3} )} \times ( 3 + x ^{\frac {-2}{3}})$$

Answer:

$$\frac {dy}{dx} = = \frac {1}{9} \times ( x+ x^{\frac {1}{3}})^{(\frac {-2}{3} )} \times ( 3 + x ^{\frac {-2}{3}})$$