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Let 
y^(4)+5x=11.
What is the value of 
(d^(2)y)/(dx^(2)) at the point 
(2,1) ?
Give an exact number.

Let y4+5x=11 y^{4}+5 x=11 .\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 y4+5x=11 y^{4}+5 x=11 .\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. Differentiate Equation: Given the equation y4+5x=11y^{4} + 5x = 11, we need to find the second derivative of yy with respect to xx, d2ydx2\frac{d^{2}y}{dx^{2}}, at the point (2,1)(2,1). To do this, we first need to differentiate both sides of the equation with respect to xx to find the first derivative dydx\frac{dy}{dx}. Differentiate y4+5x=11y^{4} + 5x = 11 with respect to xx: ddx(y4)+ddx(5x)=ddx(11)\frac{d}{dx}(y^{4}) + \frac{d}{dx}(5x) = \frac{d}{dx}(11) Using the power rule and the constant rule for differentiation, we get: yy00
  2. Solve for dydx\frac{dy}{dx}: Now we need to solve for dydx\frac{dy}{dx}:4y3dydx+5=04y^{3} \cdot \frac{dy}{dx} + 5 = 0 Subtract 55 from both sides:4y3dydx=54y^{3} \cdot \frac{dy}{dx} = -5 Divide both sides by 4y34y^{3}:dydx=54y3\frac{dy}{dx} = \frac{-5}{4y^{3}}
  3. Find Second Derivative: Next, we need to find the second derivative (d2y)/(dx2)(d^{2}y)/(dx^{2}). To do this, we differentiate (dy/dx)(dy/dx) with respect to xx again. However, since dy/dxdy/dx is a function of yy, and yy is a function of xx, we need to use the chain rule for differentiation.\newlineDifferentiate (dy/dx)=5/(4y3)(dy/dx) = -5 / (4y^{3}) with respect to xx:\newline(d2y)/(dx2)=d/dx(5/(4y3))(d^{2}y)/(dx^{2}) = d/dx(-5 / (4y^{3}))\newlineUsing the chain rule and the power rule, we get:\newline(dy/dx)(dy/dx)00\newline(dy/dx)(dy/dx)11\newline(dy/dx)(dy/dx)22
  4. Substitute dydx\frac{dy}{dx}: Now we need to substitute the expression we found for dydx\frac{dy}{dx} into the equation for the second derivative:\newlined2ydx2=15×(14y4)×(54y3)\frac{d^{2}y}{dx^{2}} = 15 \times \left(\frac{1}{4y^{4}}\right) \times \left(-\frac{5}{4y^{3}}\right)\newlineSimplify the expression:\newlined2ydx2=15×(5)/(16y7)\frac{d^{2}y}{dx^{2}} = 15 \times (-5) / (16y^{7})\newlined2ydx2=7516y7\frac{d^{2}y}{dx^{2}} = -\frac{75}{16y^{7}}
  5. Evaluate at (2,1)(2,1): Finally, we need to evaluate the second derivative at the point (2,1)(2,1). Since y=1y = 1 at this point, we substitute y=1y = 1 into the expression for the second derivative:\newlined2ydx2\frac{d^{2}y}{dx^{2}} at (2,1)(2,1) = 7516(1)7-\frac{75}{16 \cdot (1)^{7}}\newlineSimplify the expression:\newlined2ydx2\frac{d^{2}y}{dx^{2}} at (2,1)(2,1) = 7516-\frac{75}{16}

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