Question
Sea \( g(x)=\sqrt{x} \sin (x) \) \( g^{\prime}(x)=\square \)
Ask by Brooks Ayala.
Mar 07,2025 07:11
UpStudy AI Solution
Tutor-Verified Answer
Answer
La derivada de \( g(x) = \sqrt{x} \sin(x) \) es:
\[
g'(x) = \frac{1}{2\sqrt{x}} \sin(x) + \sqrt{x} \cos(x)
\]
Solution
1. Identificamos la función como el producto de dos funciones:
- \( u(x)=\sqrt{x}=x^{\frac{1}{2}} \)
- \( v(x)=\sin(x) \)
2. Aplicamos la regla del producto:
\[
g'(x)=u'(x)v(x)+u(x)v'(x)
\]
3. Calculamos \( u'(x) \):
\[
u'(x)=\frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}
\]
4. Calculamos \( v'(x) \):
\[
v'(x)=\frac{d}{dx}\left(\sin(x)\right)=\cos(x)
\]
5. Sustituimos en la fórmula del producto:
\[
g'(x)=\frac{1}{2\sqrt{x}}\sin(x)+\sqrt{x}\cos(x)
\]
La derivada de \( g(x) \) es:
\[
g'(x)=\frac{1}{2\sqrt{x}}\sin(x)+\sqrt{x}\cos(x)
\]
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Simplify this solution Extra Insights
To find the derivative of \( g(x) = \sqrt{x} \sin(x) \), we can use the product rule. The product rule states that if you have two functions multiplied together, say \( u(x) \) and \( v(x) \), then the derivative \( (uv)' \) is \( u'v + uv' \). For our function, let \( u(x) = \sqrt{x} \) and \( v(x) = \sin(x) \).
Calculating the derivatives:
- \( u'(x) = \frac{1}{2\sqrt{x}} \)
- \( v'(x) = \cos(x) \)
Now, applying the product rule:
\[
g'(x) = u'(x)v(x) + u(x)v'(x) = \frac{1}{2\sqrt{x}} \sin(x) + \sqrt{x} \cos(x)
\]
Thus, we arrive at the final expression:
\[
g'(x) = \frac{1}{2\sqrt{x}} \sin(x) + \sqrt{x} \cos(x)
\]
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