Question

$y=\frac{(e^{sinx}-1)^4+((cosx-1)e^{sinx})^3+(e^x+e^{-x})^3+x^3+2x^2+4}{e^x}$

$y''(0)=?$

Answer 1

40

Answer 2

The function is given by $y(x) = \frac{(e^{\sin x}-1)^4+((cosx-1)e^{\sin x})^3+(e^x+e^{-x})^3+x^3+2x^2+4}{e^x}$.

Let the numerator be $N(x) = (e^{\sin x}-1)^4+((cosx-1)e^{\sin x})^3+(e^x+e^{-x})^3+x^3+2x^2+4$. Then $y(x) = N(x)e^{-x}$.

To find $y''(0)$, we first find the derivatives of $y(x)$: $y'(x) = N'(x)e^{-x} - N(x)e^{-x} = (N'(x) - N(x))e^{-x}$ $y''(x) = (N''(x) - N'(x))e^{-x} - (N'(x) - N(x))e^{-x} = (N''(x) - 2N'(x) + N(x))e^{-x}$ At $x=0$, $y''(0) = (N''(0) - 2N'(0) + N(0))e^{-0} = N''(0) - 2N'(0) + N(0)$.

Let's evaluate $N(0)$, $N'(0)$, and $N''(0)$. $N(x) = (e^{\sin x}-1)^4+((cosx-1)e^{\sin x})^3+(e^x+e^{-x})^3+x^3+2x^2+4$ At $x=0$: $\sin 0 = 0$, $\cos 0 = 1$. $N(0) = (e^0-1)^4 + ((1-1)e^0)^3 + (e^0+e^{-0})^3 + 0^3 + 2(0)^2 + 4$ $N(0) = (1-1)^4 + (0 \cdot 1)^3 + (1+1)^3 + 0 + 0 + 4$ $N(0) = 0^4 + 0^3 + 2^3 + 4 = 0 + 0 + 8 + 4 = 12$.

To find $N'(0)$ and $N''(0)$, we can use Taylor series expansions of the terms around $x=0$. Recall the series expansions: $\sin x = x - \frac{x^3}{6} + O(x^5)$ $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$ $e^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + \frac{u^4}{24} + ...$

Term 1: $T_1 = (e^{\sin x}-1)^4$ $e^{\sin x} = 1 + \sin x + \frac{\sin^2 x}{2} + \frac{\sin^3 x}{6} + \frac{\sin^4 x}{24} + O(x^5)$ $= 1 + (x - \frac{x^3}{6}) + \frac{(x - \frac{x^3}{6})^2}{2} + \frac{(x)^3}{6} + \frac{(x)^4}{24} + O(x^5)$ $= 1 + x - \frac{x^3}{6} + \frac{x^2 - x^4/3 + O(x^6)}{2} + \frac{x^3}{6} + \frac{x^4}{24} + O(x^5)$ $= 1 + x + \frac{x^2}{2} + x^3(-\frac{1}{6}+\frac{1}{6}) + x^4(-\frac{1}{6}+\frac{1}{24}) + O(x^5)$ $= 1 + x + \frac{x^2}{2} - \frac{x^4}{8} + O(x^5)$ $e^{\sin x}-1 = x + \frac{x^2}{2} - \frac{x^4}{8} + O(x^5)$ $T_1 = (e^{\sin x}-1)^4 = (x + \frac{x^2}{2} - \frac{x^4}{8} + O(x^5))^4 = x^4 + 4x^3(\frac{x^2}{2}) + O(x^6) = x^4 + 2x^5 + O(x^6)$. $T_1(x) = c_4 x^4 + c_5 x^5 + ...$ $T_1'(x) = 4c_4 x^3 + 5c_5 x^4 + ... \implies T_1'(0) = 0$. $T_1''(x) = 12c_4 x^2 + 20c_5 x^3 + ... \implies T_1''(0) = 0$.

Term 2: $T_2 = ((cosx-1)e^{\sin x})^3$ $\cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$ $e^{\sin x} = 1 + x + \frac{x^2}{2} + O(x^3)$ $(cosx-1)e^{\sin x} = (-\frac{x^2}{2} + O(x^4))(1 + x + O(x^2)) = -\frac{x^2}{2} - \frac{x^3}{2} + O(x^4)$. $T_2 = ((cosx-1)e^{\sin x})^3 = (-\frac{x^2}{2} - \frac{x^3}{2} + O(x^4))^3 = (-\frac{x^2}{2})^3 + 3(-\frac{x^2}{2})^2(-\frac{x^3}{2}) + O(x^8)$ $= -\frac{x^6}{8} - \frac{3x^7}{8} + O(x^8)$. $T_2(x) = d_6 x^6 + d_7 x^7 + ...$ $T_2'(x) = 6d_6 x^5 + 7d_7 x^6 + ... \implies T_2'(0) = 0$. $T_2''(x) = 30d_6 x^4 + 42d_7 x^5 + ... \implies T_2''(0) = 0$.

Term 3: $T_3 = (e^x+e^{-x})^3$ $e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24} + O(x^5)$ $e^{-x} = 1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24} + O(x^5)$ $e^x+e^{-x} = 2 + x^2 + \frac{x^4}{12} + O(x^6)$ $T_3 = (2 + x^2 + \frac{x^4}{12} + O(x^6))^3$ $= 2^3 + 3 \cdot 2^2 (x^2 + \frac{x^4}{12} + O(x^6)) + 3 \cdot 2 (x^2 + O(x^4))^2 + (x^2 + O(x^4))^3$ $= 8 + 12(x^2 + \frac{x^4}{12} + O(x^6)) + 6(x^4 + O(x^6)) + O(x^6)$ $= 8 + 12x^2 + x^4 + 6x^4 + O(x^6) = 8 + 12x^2 + 7x^4 + O(x^6)$. $T_3(x) = 8 + 12x^2 + 7x^4 + ...$ $T_3'(x) = 24x + 28x^3 + ... \implies T_3'(0) = 0$. $T_3''(x) = 24 + 84x^2 + ... \implies T_3''(0) = 24$.

Term 4: $T_4 = x^3+2x^2+4$ $T_4'(x) = 3x^2+4x \implies T_4'(0) = 0$. $T_4''(x) = 6x+4 \implies T_4''(0) = 4$.

$N(x) = T_1 + T_2 + T_3 + T_4$. $N'(x) = T_1' + T_2' + T_3' + T_4'$. $N'(0) = T_1'(0) + T_2'(0) + T_3'(0) + T_4'(0) = 0 + 0 + 0 + 0 = 0$.

$N''(x) = T_1'' + T_2'' + T_3'' + T_4''$. $N''(0) = T_1''(0) + T_2''(0) + T_3''(0) + T_4''(0) = 0 + 0 + 24 + 4 = 28$.

Finally, $y''(0) = N''(0) - 2N'(0) + N(0)$. $y''(0) = 28 - 2(0) + 12 = 28 + 12 = 40$.

The final answer is $\boxed{40}$.

Explanation of the solution:

1. Define the numerator of the function as $N(x)$ and write $y(x) = N(x)e^{-x}$.
2. Calculate the second derivative $y''(x)$ using the product rule. Evaluate $y''(0)$ in terms of $N(0)$, $N'(0)$, and $N''(0)$.
3. Evaluate $N(0)$ by substituting $x=0$ into the expression for $N(x)$.
4. Evaluate $N'(0)$ and $N''(0)$ by finding the first and second derivatives of each term in $N(x)$ and evaluating them at $x=0$. Taylor series expansions around $x=0$ are used to efficiently determine the values of the derivatives at $x=0$. For terms starting with powers of $x$ greater than 1, the first and second derivatives at $x=0$ are zero.
5. Substitute the calculated values of $N(0)$, $N'(0)$, and $N''(0)$ into the expression for $y''(0)$.

The final answer is $\boxed{40}$.