find the derivative of the given function.\n|y|$\frac{dy}{dx}$|\n|$e^{-6x}$|$-6e^{-6x}$|\n|$e^{sqrt{x}}$||\n|…

Explanation:

Step1: Recall chain - rule for $y = e^{\sqrt{x}}$

The chain - rule states that if $y = e^{u}$ and $u=\sqrt{x}=x^{\frac{1}{2}}$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. The derivative of $y = e^{u}$ with respect to $u$ is $e^{u}$, and the derivative of $u = x^{\frac{1}{2}}$ with respect to $x$ is $\frac{1}{2}x^{-\frac{1}{2}}$. So $\frac{dy}{dx}=e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}=\frac{e^{\sqrt{x}}}{2\sqrt{x}}$.

Step2: Recall chain - rule for $y = e^{x^{7}}$

Let $u = x^{7}$. Then $y = e^{u}$. The derivative of $y$ with respect to $u$ is $e^{u}$, and the derivative of $u$ with respect to $x$ is $7x^{6}$. By the chain - rule, $\frac{dy}{dx}=e^{x^{7}}\cdot7x^{6}=7x^{6}e^{x^{7}}$.

Step3: Recall product - rule for $y = x^{7}e^{x}$

The product - rule states that if $y = uv$ where $u = x^{7}$ and $v = e^{x}$, then $\frac{dy}{dx}=u'v + uv'$. The derivative of $u=x^{7}$ is $u' = 7x^{6}$, and the derivative of $v = e^{x}$ is $v'=e^{x}$. So $\frac{dy}{dx}=7x^{6}e^{x}+x^{7}e^{x}=x^{6}e^{x}(7 + x)$.

Answer:

$y = e^{\sqrt{x}}$, $\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{2\sqrt{x}}$ $y = e^{x^{7}}$, $\frac{dy}{dx}=7x^{6}e^{x^{7}}$ $y = x^{7}e^{x}$, $\frac{dy}{dx}=x^{6}e^{x}(7 + x)$