1. **State the problem:** Find the derivative of the function $$y=\frac{2x^4 \tan x}{e^{2x}\sin x}$$ and simplify. 2. **Rewrite the function:** It helps to write the function as a quotient to apply the quotient rule: $$y=\frac{u}{v}\quad \text{where } u=2x^4 \tan x, \quad v=e^{2x} \sin x$$ 3. **Recall the quotient rule:** The derivative is $$y' = \frac{u'v - uv'}{v^2}$$ 4. **Compute** $u'$: - Derivative of $2x^4$ is $8x^3$ - Derivative of $\tan x$ is $\sec^2 x$ Using the product rule on $u=2x^4\tan x$: $$u' = (8x^3)(\tan x) + (2x^4)(\sec^2 x)$$ 5. **Compute** $v'$: - Derivative of $e^{2x}$ is $2 e^{2x}$ - Derivative of $\sin x$ is $\cos x$ Using the product rule on $v=e^{2x}\sin x$: $$v' = (2 e^{2x})(\sin x) + (e^{2x})(\cos x) = e^{2x}(2 \sin x + \cos x)$$ 6. **Substitute into quotient rule:** $$y' = \frac{\left[(8x^3 \tan x + 2x^4 \sec^2 x) \cdot e^{2x} \sin x \right] - \left[2x^4 \tan x \cdot e^{2x} (2 \sin x + \cos x) \right]}{(e^{2x} \sin x)^2}$$ 7. **Factor $e^{2x}$ from numerator terms:** $$y' = \frac{e^{2x} \left[ (8x^3 \tan x + 2x^4 \sec^2 x) \sin x - 2x^4 \tan x (2 \sin x + \cos x) \right]}{e^{4x} \sin^2 x}$$ Simplify denominator: $$= \frac{\text{Numerator}}{e^{4x} \sin^2 x}$$ 8. **Cancel $e^{2x}$ in numerator with denominator:** $$y' = \frac{(8x^3 \tan x + 2x^4 \sec^2 x) \sin x - 2x^4 \tan x (2 \sin x + \cos x)}{e^{2x} \sin^2 x}$$ 9. **Expand numerator terms:** $$= \frac{8x^3 \tan x \sin x + 2x^4 \sec^2 x \sin x - 4x^4 \tan x \sin x - 2x^4 \tan x \cos x}{e^{2x} \sin^2 x}$$ 10. **Simplify terms inside numerator:** Recall $\tan x = \frac{\sin x}{\cos x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$. - $8x^3 \tan x \sin x = 8x^3 \frac{\sin x}{\cos x} \sin x = 8x^3 \frac{\sin^2 x}{\cos x}$ - $2x^4 \sec^2 x \sin x = 2x^4 \frac{1}{\cos^2 x} \sin x$ - $4x^4 \tan x \sin x = 4x^4 \frac{\sin x}{\cos x} \sin x = 4x^4 \frac{\sin^2 x}{\cos x}$ - $2x^4 \tan x \cos x = 2x^4 \frac{\sin x}{\cos x} \cos x = 2x^4 \sin x$ So numerator becomes: $$8x^3 \frac{\sin^2 x}{\cos x} + 2x^4 \frac{\sin x}{\cos^2 x} - 4x^4 \frac{\sin^2 x}{\cos x} - 2x^4 \sin x$$ 11. **Group terms over common denominators:** Multiply numerator and denominator terms by $\cos^2 x$ to combine: Express all terms with denominator $\cos^2 x$: - $8x^3 \frac{\sin^2 x}{\cos x} = 8x^3 \sin^2 x \frac{\cos x}{\cos^2 x} = 8x^3 \sin^2 x \frac{1}{\cos x}$ (already as is; keep as $8x^3 \sin^2 x / \cos x$) Better to write as: $$= \frac{8x^3 \sin^2 x \cos x + 2x^4 \sin x - 4x^4 \sin^2 x \cos x - 2x^4 \sin x \cos^2 x}{\cos^2 x}$$ (After multiplying numerator and denominator by $\cos^2 x$) 12. **Simplify numerator: grouping like terms:** Group the $x^4 \sin x$ terms: $$2x^4 \sin x - 2x^4 \sin x \cos^2 x = 2x^4 \sin x (1 - \cos^2 x) = 2x^4 \sin x \sin^2 x$$ (using $\sin^2 x + \cos^2 x=1$) So numerator: $$8x^3 \sin^2 x \cos x - 4x^4 \sin^2 x \cos x + 2x^4 \sin^3 x$$ 13. **Factor terms where possible:** Factor $\sin^2 x \cos x$ from first two terms: $$\sin^2 x \cos x (8x^3 - 4x^4) + 2x^4 \sin^3 x = 4x^3 \sin^2 x \cos x (2 - x) + 2x^4 \sin^3 x$$ 14. **Final derivative:** $$y' = \frac{4x^3 \sin^2 x \cos x (2 - x) + 2x^4 \sin^3 x}{e^{2x} \sin^2 x \cos^2 x}$$ After cancelling one $\sin^2 x$ if desired: $$y' = \frac{4x^3 \cos x (2 - x) + 2x^4 \sin x}{e^{2x} \cos^2 x \sin x}$$ This is a simplified form. **Final answer:** $$\boxed{y' = \frac{4x^3 \cos x (2 - x) + 2x^4 \sin x}{e^{2x} \cos^2 x \sin x}}$$