find the derivative of $y = 3^{sqrt{t}}$ with respect to $t$.\n$\frac{dy}{dt}=square$

Explanation:

Step1: Rewrite the function

Let $y = 3^{\sqrt{t}}$. We use the formula $a^u=e^{u\ln a}$, so $y = e^{\sqrt{t}\ln 3}$. Let $u = \sqrt{t}=t^{\frac{1}{2}}$, then $y = e^{u\ln 3}$.

Step2: Apply the chain - rule

The chain - rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$. Since $y = e^{u\ln 3}$, then $\frac{dy}{du}=\ln 3\cdot e^{u\ln 3}$ (because the derivative of $e^{ax}$ with respect to $x$ is $ae^{ax}$, here $a = \ln 3$). Second, find $\frac{du}{dt}$. Since $u=t^{\frac{1}{2}}$, then $\frac{du}{dt}=\frac{1}{2}t^{-\frac{1}{2}}$ by the power - rule ($\frac{d}{dt}(t^n)=nt^{n - 1}$).

Step3: Substitute $u$ back and calculate $\frac{dy}{dt}$

Substitute $u = \sqrt{t}$ back into $\frac{dy}{du}$ and multiply by $\frac{du}{dt}$. We have $\frac{dy}{dt}=\ln 3\cdot e^{\sqrt{t}\ln 3}\cdot\frac{1}{2\sqrt{t}}$. Since $e^{\sqrt{t}\ln 3}=3^{\sqrt{t}}$, then $\frac{dy}{dt}=\frac{\ln 3\cdot3^{\sqrt{t}}}{2\sqrt{t}}$.

Answer:

$\frac{\ln 3\cdot3^{\sqrt{t}}}{2\sqrt{t}}$