Question

find the derivative of the function.

\\q = \sqrt5{x^4 + 6x}\\

\\q = \\

Explanation:

Rewrite the function using rational exponents

To find the derivative of \(q = \sqrt[5]{x^4 + 6x}\), we first express the radical as a fractional power:

$$q = (x^4 + 6x)^{\frac{1}{5}}$$

Apply the chain rule

We use the Chain Rule to differentiate a composite function \(u(x)^n\):

$$\frac{dq}{dx} = n \cdot u(x)^{n-1} \cdot u'(x)$$

Here, \(u(x) = x^4 + 6x\) and \(n = \frac{1}{5}\).

Differentiate the outer and inner functions

The derivative of the outer function with respect to \(u\) is:

$$\frac{1}{5}(x^4 + 6x)^{-\frac{4}{5}}$$

The derivative of the inner function \(u(x) = x^4 + 6x\) with respect to \(x\) is:

$$u'(x) = 4x^3 + 6$$

Combine the derivatives

Multiplying the outer and inner derivatives yields:

$$q' = \frac{1}{5}(x^4 + 6x)^{-\frac{4}{5}} \cdot (4x^3 + 6)$$

Simplify the expression

We rewrite the negative exponent in the denominator and simplify:

$$q' = \frac{4x^3 + 6}{5(x^4 + 6x)^{\frac{4}{5}}} = \frac{4x^3 + 6}{5\sqrt[5]{(x^4 + 6x)^4}}$$

Answer:

Find the derivative of the function.
\(q = \sqrt[5]{x^4 + 6x}\)

\(q' =\) <blank>\(\frac{4x^3 + 6}{5\sqrt[5]{(x^4 + 6x)^4}}\)</blank>