Math  /  Calculus

QuestionExercies find the second derivative of y=f(x)y=f(x) at x=0x=0 where yy is determined by y5+2yx3x7=0y^{5}+2 y-x-3 x^{7}=0

Studdy Solution
Evaluate d2ydx2 \frac{d^2y}{dx^2} at x=0 x = 0 .
First, find y y when x=0 x = 0 from the original equation:
y5+2y=0y^5 + 2y = 0
Factor out y y :
y(y4+2)=0y(y^4 + 2) = 0
Thus, y=0 y = 0 (since y4+20 y^4 + 2 \neq 0 ).
Substitute x=0 x = 0 and y=0 y = 0 into the expression for d2ydx2 \frac{d^2y}{dx^2} :
d2ydx2=(5(0)4+2)(126(0)5)(1+21(0)6)(20(0)312)(5(0)4+2)2\frac{d^2y}{dx^2} = \frac{(5(0)^4 + 2)(126(0)^5) - (1 + 21(0)^6)(20(0)^3 \cdot \frac{1}{2})}{(5(0)^4 + 2)^2}
Simplify:
d2ydx2=004=0\frac{d^2y}{dx^2} = \frac{0 - 0}{4} = 0
The second derivative at x=0 x = 0 is 0 \boxed{0} .

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