Question
QUESTION 3 (16 manks) Find the derivative for the following functions, giving your answers in their sinplast form. a. \( y=4 x^{3}-2 \sqrt{x}-\frac{1}{x^{3}} \)
Ask by Dickson Pena. in Singapore
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The derivative of \( y = 4x^{3} - 2\sqrt{x} - \frac{1}{x^{3}} \) is \( \frac{dy}{dx} = 12x^{2} - \frac{1}{\sqrt{x}} + \frac{3}{x^{4}} \).
Solution
To find the derivative of the function \( y = 4x^{3} - 2\sqrt{x} - \frac{1}{x^{3}} \), we'll differentiate each term separately and then combine the results.
1. **Differentiate \( 4x^{3} \):**
\[
\frac{d}{dx}(4x^{3}) = 12x^{2}
\]
2. **Differentiate \( -2\sqrt{x} \):**
\[
-2\sqrt{x} = -2x^{1/2}
\]
\[
\frac{d}{dx}(-2x^{1/2}) = -2 \cdot \frac{1}{2}x^{-1/2} = -x^{-1/2} = -\frac{1}{\sqrt{x}}
\]
3. **Differentiate \( -\frac{1}{x^{3}} \):**
\[
-\frac{1}{x^{3}} = -x^{-3}
\]
\[
\frac{d}{dx}(-x^{-3}) = -(-3)x^{-4} = 3x^{-4} = \frac{3}{x^{4}}
\]
**Combine all the derivatives:**
\[
\frac{dy}{dx} = 12x^{2} - \frac{1}{\sqrt{x}} + \frac{3}{x^{4}}
\]
**Final Answer:**
\[
\frac{dy}{dx} = 12\,x^{2} - \frac{1}{\,\sqrt{x\,}} + \frac{3}{\,x^{4}\,}
\]
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Simplify this solution The Deep Dive
To find the derivative of \( y=4 x^{3}-2 \sqrt{x}-\frac{1}{x^{3}} \), we apply the power rule, the derivative of a square root, and the derivative of \( x^{-n} \). 1. The derivative of \( 4 x^{3} \) is \( 12 x^{2} \). 2. The derivative of \( -2 \sqrt{x} \) is \( -\frac{2}{2\sqrt{x}} = -\frac{1}{\sqrt{x}} \) or simplified, \( -x^{-\frac{1}{2}} \). 3. The derivative of \( -\frac{1}{x^{3}} \) is \( 3x^{-4} \). Now put it all together: \[ \frac{dy}{dx} = 12 x^{2} - \frac{1}{\sqrt{x}} + 3x^{-4} \] Or in simplest form: \[ \frac{dy}{dx} = 12 x^{2} - x^{-\frac{1}{2}} + \frac{3}{x^{4}} \]
