Rewrite the function using negative exponents to make differentiation easier.

Question

ScanSolve AI Tutor · Accepted Answer

Okay, Ni_Fe_Mi✨. Let's differentiate this function.

The function to differentiate is $y = \left(2x^2 - \frac{3}{x^2}\right)^{1/3}$ .

Rewrite the function using negative exponents to make differentiation easier.
$y = (2x^2 - 3x^{-2})^{1/3}$
Apply the chain rule.
The chain rule states that if $y = f(g(x))$ , then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$ . Here, $f(u) = u^{1/3}$ and $g(x) = 2x^2 - 3x^{-2}$ .
$\frac{dy}{dx} = \frac{1}{3}(2x^2 - 3x^{-2})^{\frac{1}{3}-1} \cdot \frac{d}{dx}(2x^2 - 3x^{-2})$
Differentiate the inner function, $2x^2 - 3x^{-2}$ , using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ .
$\frac{d}{dx}(2x^2 - 3x^{-2}) = 2(2x^{2-1}) - 3(-2x^{-2-1})$ $= 4x - (-6x^{-3})$ $= 4x + 6x^{-3}$
Substitute the derivative of the inner function back into the expression from Step 2 and simplify the exponent.
$\frac{dy}{dx} = \frac{1}{3}(2x^2 - 3x^{-2})^{-2/3} (4x + 6x^{-3})$
Rewrite the expression with positive exponents and simplify the terms.
$\frac{dy}{dx} = \frac{4x + \frac{6}{x^3}}{3\left(2x^2 - \frac{3}{x^2}\right)^{2/3}}$
To simplify the numerator, find a common denominator:
$4x + \frac{6}{x^3} = \frac{4x \cdot x^3}{x^3} + \frac{6}{x^3} = \frac{4x^4 + 6}{x^3}$
So, the expression becomes:
$\frac{dy}{dx} = \frac{\frac{4x^4 + 6}{x^3}}{3\left(2x^2 - \frac{3}{x^2}\right)^{2/3}}$
You can also factor out a 2 from the numerator:
$\frac{dy}{dx} = \frac{2(2x^4 + 3)}{3x^3\left(2x^2 - \frac{3}{x^2}\right)^{2/3}}$
The derivative is $\boxed{\mathbf{\frac{2(2x^4+3)}{3x^3\left(2x^2 - \frac{3}{x^2}\right)^{2/3}}}}$ .

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