To show that if [tex]\( y = \sqrt{x + \sqrt{x}} \)[/tex], then [tex]\( \frac{dy}{dx} = \frac{1}{2y}\left(1 + \frac{1}{2\sqrt{x}}\right) \)[/tex], let's follow a step-by-step approach.
1. Define the function:
[tex]\[
y = \sqrt{x + \sqrt{x}}
\][/tex]
2. Differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
We will use the chain rule for differentiation. First, denote the inside function:
[tex]\[
u = x + \sqrt{x}
\][/tex]
So, [tex]\( y = \sqrt{u} \)[/tex].
3. Differentiate [tex]\( y \)[/tex] with respect to [tex]\( u \)[/tex]:
[tex]\[
\frac{dy}{du} = \frac{1}{2\sqrt{u}}
\][/tex]
4. Differentiate [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
u = x + \sqrt{x}
\][/tex]
[tex]\[
\frac{du}{dx} = 1 + \frac{1}{2\sqrt{x}}
\][/tex]
5. Apply the chain rule:
[tex]\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
\][/tex]
Substitute the results from steps 3 and 4:
[tex]\[
\frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \left( 1 + \frac{1}{2\sqrt{x}} \right)
\][/tex]
6. Substitute back [tex]\( u \)[/tex] with [tex]\( x + \sqrt{x} \)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{1}{2\sqrt{x + \sqrt{x}}} \left( 1 + \frac{1}{2\sqrt{x}} \right)
\][/tex]
7. Recognize that [tex]\( y = \sqrt{x + \sqrt{x}} \)[/tex]:
Thus, we can substitute [tex]\( y \)[/tex] directly back into the expression:
[tex]\[
\frac{dy}{dx} = \frac{1}{2y} \left( 1 + \frac{1}{2\sqrt{x}} \right)
\][/tex]
Therefore, we have successfully shown that:
[tex]\[
\frac{dy}{dx} = \frac{1}{2y} \left( 1 + \frac{1}{2\sqrt{x}} \right)
\][/tex]
