find the derivative of the function. y = √(11x + √(11x + √(11x))) y =

Explanation:

Step1: Let $u = 11x+\sqrt{11x + \sqrt{11x}}$

$y=\sqrt{u}=u^{\frac{1}{2}}$

Step2: Apply the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$

Step3: Let $v = 11x+\sqrt{11x}$

$u = 11x + \sqrt{v}$

Step4: Find $\frac{du}{dx}$

$\frac{du}{dx}=11+\frac{1}{2\sqrt{v}}\cdot\frac{dv}{dx}$

Step5: Let $w = 11x$

$v = 11x+\sqrt{w}$

Step6: Find $\frac{dv}{dx}$

$\frac{dv}{dx}=11+\frac{1}{2\sqrt{w}}\cdot\frac{dw}{dx}$

Step7: Since $\frac{dw}{dx}=11$

$\frac{dv}{dx}=11+\frac{11}{2\sqrt{11x}}$

Step8: Substitute $\frac{dv}{dx}$ into $\frac{du}{dx}$

$\frac{du}{dx}=11+\frac{1}{2\sqrt{11x+\sqrt{11x}}}\left(11 + \frac{11}{2\sqrt{11x}}\right)$

Step9: Substitute $\frac{du}{dx}$ and $\frac{dy}{du}$ into $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

$\frac{dy}{dx}=\frac{1}{2\sqrt{11x+\sqrt{11x+\sqrt{11x}}}}\left(11+\frac{11}{2\sqrt{11x+\sqrt{11x}}}\left(1 + \frac{1}{2\sqrt{11x}}\right)\right)$

Answer:

$\frac{11}{2\sqrt{11x+\sqrt{11x+\sqrt{11x}}}}\left(1+\frac{1}{2\sqrt{11x+\sqrt{11x}}}\left(1 + \frac{1}{2\sqrt{11x}}\right)\right)$