Find the derivative of the function.

[tex]\[
\begin{array}{l}
y = 4 \sqrt{x} + 5 x^{\frac{1}{5}} \\
\frac{d y}{d x} = \square
\end{array}
\][/tex]



Answer :

Sure, let's find the derivative of the given function step-by-step.

We start with the function:
[tex]\[ y = 4 \sqrt{x} + 5 x^{\frac{1}{5}} \][/tex]

To differentiate this function, we will apply the power rule for derivatives, which states that for any term [tex]\( x^n \)[/tex], the derivative is [tex]\( n x^{n-1} \)[/tex].

First, let's rewrite the square root of [tex]\( x \)[/tex] using exponents:
[tex]\[ \sqrt{x} = x^{\frac{1}{2}} \][/tex]

Now the function becomes:
[tex]\[ y = 4 x^{\frac{1}{2}} + 5 x^{\frac{1}{5}} \][/tex]

Next, we differentiate each term individually.

1. For the first term, [tex]\( 4 x^{\frac{1}{2}} \)[/tex]:

[tex]\[ \frac{d}{dx} \left( 4 x^{\frac{1}{2}} \right) = 4 \cdot \frac{1}{2} x^{\frac{1}{2} - 1} = 2 x^{-\frac{1}{2}} \][/tex]

2. For the second term, [tex]\( 5 x^{\frac{1}{5}} \)[/tex]:

[tex]\[ \frac{d}{dx} \left( 5 x^{\frac{1}{5}} \right) = 5 \cdot \frac{1}{5} x^{\frac{1}{5} - 1} = x^{-\frac{4}{5}} \][/tex]

So the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{d y}{d x} = 2 x^{-\frac{1}{2}} + x^{-\frac{4}{5}} \][/tex]

Let's rewrite those terms to make them easier to understand. Remembering that:

[tex]\[ x^{-\frac{1}{2}} = \frac{1}{x^{\frac{1}{2}}} = \frac{1}{\sqrt{x}} \][/tex]

[tex]\[ x^{-\frac{4}{5}} = \frac{1}{x^{\frac{4}{5}}} \][/tex]

And since [tex]\(\frac{1}{x^{\frac{4}{5}}} = \frac{1}{x^{0.8}} \)[/tex], then:

[tex]\[ \frac{d y}{d x} = \frac{2}{\sqrt{x}} + \frac{1}{x^{0.8}} \][/tex]

Thus, the final result is:
[tex]\[ \boxed{\frac{d y}{d x} = \frac{1}{x^{0.8}} + \frac{2}{\sqrt{x}}} \][/tex]