Find the derivative of the function. $$ y=x^{-4}+x^{-2}-8 $$ $$ y^{\prime}(x)=\square $$

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To find the derivative of the function $$ y = x^{-4} + x^{-2} - 8 $$, we'll differentiate each term separately using the power rule. The power rule states that if $$ y = x^n $$, then $$ y' = n \cdot x^{n-1} $$.

1. **Differentiate $$ x^{-4} $$:**
   $$
   \frac{d}{dx}(x^{-4}) = -4x^{-5}
   $$

2. **Differentiate $$ x^{-2} $$:**
   $$
   \frac{d}{dx}(x^{-2}) = -2x^{-3}
   $$

3. **Differentiate the constant $$-8$$:**
   $$
   \frac{d}{dx}(-8) = 0
   $$

Combining these results, the derivative $$ y'(x) $$ is:
$$
y'(x) = -4x^{-5} - 2x^{-3}
$$

Alternatively, you can write it with positive exponents in the denominator:
$$
y'(x) = -\frac{4}{x^5} - \frac{2}{x^3}
$$

**Final Answer:**
$$
y^{\prime}(x) = -4\,x^{-5} - 2\,x^{-3}
$$
The derivative of the function $$ y = x^{-4} + x^{-2} - 8 $$ is $$ y^{\prime}(x) = -4x^{-5} - 2x^{-3} $$.

Find the derivative of the function. \[ y=x^{-4}+x^{-2}-8 \] \( y^{\prime}(x)=\square \)

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