Using the product rule, $$ \frac{d y}{d x}=v \frac{d u}{d x}+u \frac{d v}{d x} $$ the derivative of $$ y=\left(-10 x^{2}-7 \sqrt[5]{x^{2}}+9\right)\left(2 x^{3}+4\right) $$ is $$ \frac{d y}{d x}=\left(-20 x-\frac{14}{5 \sqrt[5]{x^{3}}}\right)\left(2 x^{3}+4\right)+6 x^{2}\left(-10 x^{2}-7 \sqrt[5]{x^{2}}+9\right) $$ Is the derivative TRUE or FALSE?

Question

UpStudy AI · Accepted Answer

**Step 1. Identify the functions**

Let  
$$
u(x) = -10x^2 - 7\sqrt[5]{x^2} + 9 = -10x^2 - 7x^{\frac{2}{5}}+ 9,
$$
$$
v(x) = 2x^3 + 4.
$$

**Step 2. Compute the derivatives**

Differentiate $$ u(x) $$:
$$
\frac{d}{dx}(-10x^2) = -20x,
$$
$$
\frac{d}{dx}\left(-7x^{\frac{2}{5}}\right) = -7 \cdot \frac{2}{5}x^{\frac{2}{5}-1} = -\frac{14}{5}x^{-\frac{3}{5}},
$$
$$
\frac{d}{dx}(9) = 0.
$$
Thus,
$$
u'(x) = -20x - \frac{14}{5}x^{-\frac{3}{5}}.
$$

Differentiate $$ v(x) $$:
$$
v'(x) = \frac{d}{dx}(2x^3) + \frac{d}{dx}(4) = 6x^2 + 0 = 6x^2.
$$

**Step 3. Apply the Product Rule**

The product rule states that
$$
\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x).
$$
Substitute the computed functions and derivatives:
$$
\frac{dy}{dx} = \left(-20x - \frac{14}{5}x^{-\frac{3}{5}}\right)(2x^3 + 4) + \left(-10x^2 - 7x^{\frac{2}{5}} + 9\right)(6x^2).
$$

**Step 4. Verify the Given Answer**

The provided expression is:
$$
\frac{dy}{dx} = \left(-20 x-\frac{14}{5 \sqrt[5]{x^{3}}}\right)\left(2 x^{3}+4\right)+ 6 x^{2}\left(-10 x^{2}-7 \sqrt[5]{x^{2}}+9\right).
$$

Notice that:
- $$ \sqrt[5]{x^2} = x^{\frac{2}{5}} $$
- $$ \sqrt[5]{x^3} = x^{\frac{3}{5}} $$
- $$ x^{-\frac{3}{5}} = \frac{1}{x^{\frac{3}{5}}} $$

Thus, the expression matches exactly the derivative we derived.

**Conclusion**

The derivative provided is TRUE.
The derivative is TRUE.

Using the product rule,
the derivative of is

Is the derivative TRUE or FALSE?

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Using the product rule, the derivative of is Is the derivative TRUE or FALSE?