Question

section 2.6: chain rule (homework)
score: 160/170 answered: 16/17
question 17
if $f(t)=sqrt{t + sqrt{t}}$, find $f(t)$. use exact values.
$f(t)=$

Explanation:

Step1: Let $u = t+\sqrt{t}$

$f(t)=\sqrt{u}$

Step2: Differentiate $y = \sqrt{u}=u^{\frac{1}{2}}$ with respect to $u$

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

Step3: Differentiate $u=t + t^{\frac{1}{2}}$ with respect to $t$

$\frac{du}{dt}=1+\frac{1}{2}t^{-\frac{1}{2}}$

Step4: Apply the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$

$\frac{dy}{dt}=\frac{1}{2}(t + \sqrt{t})^{-\frac{1}{2}}\cdot(1+\frac{1}{2\sqrt{t}})$

Answer:

$\frac{1+\frac{1}{2\sqrt{t}}}{2\sqrt{t+\sqrt{t}}}$