find the derivative of y = 6x^{-2sqrt{x}}. be sure to include parentheses around the arguments of any…

Explanation:

Step1: Rewrite the function

Rewrite $y = 6x^{-2\sqrt{x}}$ as $y=6e^{-2\sqrt{x}\ln(x)}$ using the formula $a^b = e^{b\ln(a)}$.

Step2: Apply the chain - rule

Let $u=-2\sqrt{x}\ln(x)$. Then $y = 6e^{u}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=6e^{u}$. Now find $\frac{du}{dx}$ using the product - rule. If $u = - 2\sqrt{x}\ln(x)=-2x^{\frac{1}{2}}\ln(x)$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=-2x^{\frac{1}{2}}$ and $v = \ln(x)$. $u^\prime=-2\times\frac{1}{2}x^{-\frac{1}{2}}=-x^{-\frac{1}{2}}$ and $v^\prime=\frac{1}{x}$. So $\frac{du}{dx}=-x^{-\frac{1}{2}}\ln(x)-2x^{\frac{1}{2}}\times\frac{1}{x}=-x^{-\frac{1}{2}}\ln(x)-2x^{-\frac{1}{2}}=-x^{-\frac{1}{2}}(\ln(x) + 2)$.

Step3: Apply the chain - rule again

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=6e^{u}$ and $\frac{du}{dx}=-x^{-\frac{1}{2}}(\ln(x)+2)$ and $u=-2\sqrt{x}\ln(x)$ back in. We get $\frac{dy}{dx}=6e^{-2\sqrt{x}\ln(x)}\left(-x^{-\frac{1}{2}}(\ln(x) + 2)\right)$. Since $6e^{-2\sqrt{x}\ln(x)}=6x^{-2\sqrt{x}}$, then $y^\prime=-6x^{-2\sqrt{x}-\frac{1}{2}}(\ln(x)+2)$.

Answer:

$-6x^{-2\sqrt{x}-\frac{1}{2}}(\ln(x)+2)$