Question

find the derivative of the function.
y = \sqrt{7x+sqrt{7x + sqrt{7x}}}
y =

Explanation:

Step1: Let $u = 7x+\sqrt{7x + \sqrt{7x}}$

The function $y=\sqrt{u}=u^{\frac{1}{2}}$. By the chain - rule, $y'=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$:
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}=\frac{1}{2\sqrt{7x+\sqrt{7x + \sqrt{7x}}}}$

Step2: Now find $\frac{du}{dx}$

Let $v = 7x+\sqrt{7x}$. Then $u = 7x+\sqrt{v}$.
First, find the derivative of $u$ with respect to $v$ and $v$ with respect to $x$.
$\frac{du}{dv}=\frac{1}{2\sqrt{v}}$, and $\frac{dv}{dx}=7+\frac{1}{2\sqrt{7x}}\cdot7$
$\frac{du}{dx}=7+\frac{1}{2\sqrt{v}}\cdot(7 + \frac{7}{2\sqrt{7x}})$
Substitute $v = 7x+\sqrt{7x}$ back in:
$\frac{du}{dx}=7+\frac{7+\frac{7}{2\sqrt{7x}}}{2\sqrt{7x+\sqrt{7x}}}$

Step3: Calculate $y'$

By the chain - rule $y'=\frac{dy}{du}\cdot\frac{du}{dx}$
$y'=\frac{7+\frac{7+\frac{7}{2\sqrt{7x}}}{2\sqrt{7x+\sqrt{7x}}}}{2\sqrt{7x+\sqrt{7x + \sqrt{7x}}}}$
Simplify the expression:
$y'=\frac{7}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}+\frac{7}{4\sqrt{(7x+\sqrt{7x+\sqrt{7x}})(7x+\sqrt{7x})}}+\frac{7}{8\sqrt{(7x+\sqrt{7x+\sqrt{7x}})(7x+\sqrt{7x})7x}}$

Answer:

$\frac{7}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}+\frac{7}{4\sqrt{(7x+\sqrt{7x+\sqrt{7x}})(7x+\sqrt{7x})}}+\frac{7}{8\sqrt{(7x+\sqrt{7x+\sqrt{7x}})(7x+\sqrt{7x})7x}}$