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Let 
y^(4)-2x=5.
What is the value of 
(d^(2)y)/(dx^(2)) at the point 
(-2,1) ?
Give an exact number.

Let y42x=5 y^{4}-2 x=5 .\newlineWhat is the value of d2ydx2 \frac{d^{2} y}{d x^{2}} at the point (2,1) (-2,1) ?\newlineGive an exact number.

Full solution

Q. Let y42x=5 y^{4}-2 x=5 .\newlineWhat is the value of d2ydx2 \frac{d^{2} y}{d x^{2}} at the point (2,1) (-2,1) ?\newlineGive an exact number.
  1. Given Equation Differentiation: We are given the equation y42x=5y^{4} - 2x = 5. To find the second derivative of yy with respect to xx, we first need to differentiate both sides of the equation with respect to xx.
    Differentiate y42x=5y^{4} - 2x = 5 with respect to xx:
    ddx(y4)ddx(2x)=ddx(5)\frac{d}{dx}(y^{4}) - \frac{d}{dx}(2x) = \frac{d}{dx}(5)
    4y3dydx2=04y^{3} \cdot \frac{dy}{dx} - 2 = 0
  2. First Derivative Calculation: Now, we need to solve for dydx\frac{dy}{dx}, which is the first derivative of yy with respect to xx.4y3dydx=24y^{3} \cdot \frac{dy}{dx} = 2dydx=24y3\frac{dy}{dx} = \frac{2}{4y^{3}}dydx=12y3\frac{dy}{dx} = \frac{1}{2y^{3}}
  3. Second Derivative Calculation: Next, we differentiate the first derivative with respect to xx to find the second derivative.ddx(dydx)=ddx(12y3)\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{1}{2y^{3}}\right)Using the chain rule and the quotient rule, we get:d2ydx2=6y2(dydx)22y3\frac{d^{2}y}{dx^{2}} = \frac{-6y^{2} \cdot \left(\frac{dy}{dx}\right)^{2}}{2y^{3}}
  4. Substitution for Given Point: We substitute the value of dydx\frac{dy}{dx} from the previous step into the expression for the second derivative.d2ydx2=6y2×(12y3)2/(2y3)\frac{d^{2}y}{dx^{2}} = -6y^{2} \times \left(\frac{1}{2y^{3}}\right)^{2} / \left(2y^{3}\right)d2ydx2=6y2×(14y6)/(2y3)\frac{d^{2}y}{dx^{2}} = -6y^{2} \times \left(\frac{1}{4y^{6}}\right) / \left(2y^{3}\right)d2ydx2=6y2/(8y9)\frac{d^{2}y}{dx^{2}} = -6y^{2} / \left(8y^{9}\right)d2ydx2=34y7\frac{d^{2}y}{dx^{2}} = -\frac{3}{4y^{7}}
  5. Final Result: Finally, we substitute the given point (2,1)(-2,1) into the expression for the second derivative.\newline(d2y)/(dx2)(d^{2}y)/(dx^{2}) at the point (2,1)(-2,1) is:\newline(d2y)/(dx2)=3/(4×17)(d^{2}y)/(dx^{2}) = -3 / (4 \times 1^{7})\newline(d2y)/(dx2)=3/4(d^{2}y)/(dx^{2}) = -3 / 4

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