Question
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Find the derivative of \( y=3^{\sqrt{x}} \) with respect to \( x \). \( \frac{d y}{d x}=\square \)

Ask by Bond French. in the United States
Mar 09,2025

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Answer

The derivative of \( y = 3^{\sqrt{x}} \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{3^{\sqrt{x}} \ln(3)}{2\sqrt{x}} \).

Solution

We start with the function \[ y = 3^{\sqrt{x}} \] recognizing that it is of the form \( y = a^{u(x)} \) where \( a = 3 \) and \( u(x) = \sqrt{x} \). **Step 1:** Differentiate using the chain rule for exponential functions. The derivative of \( a^{u(x)} \) is \[ \frac{dy}{dx} = a^{u(x)} \ln(a) \cdot u'(x) \] Substitute \( a = 3 \) and \( u(x) = \sqrt{x} \): \[ \frac{dy}{dx} = 3^{\sqrt{x}} \ln(3) \cdot \frac{d}{dx} \left( \sqrt{x} \right) \] **Step 2:** Compute the derivative of \( \sqrt{x} \). Recall that \[ \sqrt{x} = x^{\frac{1}{2}} \] Using the power rule: \[ \frac{d}{dx} \left( x^{\frac{1}{2}} \right) = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \] **Step 3:** Substitute the derivative of \( \sqrt{x} \) back into the expression. \[ \frac{dy}{dx} = 3^{\sqrt{x}} \ln(3) \cdot \frac{1}{2\sqrt{x}} \] Thus, the derivative is \[ \frac{dy}{dx} = \frac{3^{\sqrt{x}} \ln(3)}{2\sqrt{x}} \]

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To find the derivative of \( y = 3^{\sqrt{x}} \) with respect to \( x \), we can use the chain rule and properties of logarithms. First, rewrite the function using the natural exponential function: \[ y = e^{\sqrt{x} \ln(3)} \] Now, apply the chain rule: \[ \frac{dy}{dx} = e^{\sqrt{x} \ln(3)} \cdot \frac{d}{dx}(\sqrt{x} \ln(3)) \] Now compute \( \frac{d}{dx}(\sqrt{x}) \): \[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \] So, the full derivative becomes: \[ \frac{dy}{dx} = 3^{\sqrt{x}} \cdot \ln(3) \cdot \frac{1}{2\sqrt{x}} \] Thus, the derivative is: \[ \frac{dy}{dx} = \frac{3^{\sqrt{x}} \ln(3)}{2\sqrt{x}} \] So, filling in the square, we get: \[ \frac{d y}{d x} = \frac{3^{\sqrt{x}} \ln(3)}{2\sqrt{x}} \]

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