How do you differentiate  y =sqrt( x + sqrtx ) using the chain rule?

Mar 17, 2018

I get $\frac{1}{2 \sqrt{x + \sqrt{x}}} + \frac{1}{4 \sqrt{{x}^{2} + x \sqrt{x}}}$.

Explanation:

We are given: $y = \sqrt{x + \sqrt{x}}$

The chain rule states that

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

So, we let $u = x + \sqrt{x} , \therefore f = \sqrt{u}$.

Now, we differentiate both.

:.(du)/dx=1+1/(2sqrt(x)

$\frac{\mathrm{df}}{\mathrm{du}} = \frac{1}{2 \sqrt{u}}$

So, $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1}{2 \sqrt{u}} \left(1 + \frac{1}{2 \sqrt{x}}\right)$

$= \frac{1}{2 \sqrt{u}} + \frac{1}{4 \sqrt{u x}}$

Reversing the substitution that $u = x + \sqrt{x}$, we get

=1/(2sqrt(x+sqrt(x)))+1/(4sqrt(x(x+sqrt(x)))

$= \frac{1}{2 \sqrt{x + \sqrt{x}}} + \frac{1}{4 \sqrt{{x}^{2} + x \sqrt{x}}}$

Mar 17, 2018

color(blue)((1+1/(2sqrt(x)))/(2sqrt(x+sqrt(x)))

Explanation:

For the chain rule we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Re-writing $\sqrt{x + \sqrt{x}}$ as ${\left(x + {x}^{\frac{1}{2}}\right)}^{\frac{1}{2}}$

Letting: $u = \left(x + {x}^{\frac{1}{2}}\right)$

We now have:

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

Using the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(u\right)}^{- \frac{1}{2}} \cdot \left(1 + \frac{1}{2} {x}^{- \frac{1}{2}}\right)$

Substituting: $u = \left(x + {x}^{\frac{1}{2}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(x + {x}^{\frac{1}{2}}\right)}^{- \frac{1}{2}} \cdot \left(1 + \frac{1}{2} {x}^{- \frac{1}{2}}\right)$

Simplifying:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(x + {x}^{\frac{1}{2}}\right)}^{- \frac{1}{2}} \cdot \left(1 + \frac{1}{2} {x}^{- \frac{1}{2}}\right)$

color(white)(888.)=(1+1/(2x^(1/2)))/(2(x+x^(1/2))^(1/2))=(1+1/(2sqrt(x)))/(2sqrt(x+sqrt(x)))=color(blue)((1+1/(2sqrt(x)))/(2sqrt(x+sqrt(x)))