Question

let $h(x)=sqrt{xln(x)}$. find $h(x)$.

Explanation:

Step1: Rewrite the function

Rewrite $h(x)=\sqrt{x\ln(x)}$ as $h(x)=(x\ln(x))^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=x\ln(x)$, then $h^\prime(x)=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime$. First, find $u^\prime$ using the product - rule. The product - rule for $u = x\ln(x)$ where $a = x$ and $b=\ln(x)$ gives $u^\prime=a^\prime b+ab^\prime$. Since $a^\prime = 1$ and $b^\prime=\frac{1}{x}$, then $u^\prime=1\cdot\ln(x)+x\cdot\frac{1}{x}=\ln(x) + 1$.

Step3: Substitute $u$ and $u^\prime$ into the chain - rule formula

Substitute $u=x\ln(x)$ and $u^\prime=\ln(x)+1$ into $h^\prime(x)=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime$. We get $h^\prime(x)=\frac{1}{2}(x\ln(x))^{-\frac{1}{2}}\cdot(\ln(x)+1)=\frac{\ln(x) + 1}{2\sqrt{x\ln(x)}}$.

Answer:

$h^\prime(x)=\frac{\ln(x)+1}{2\sqrt{x\ln(x)}}$