Q. Let y4−2x=5.What is the value of dx2d2y at the point (−2,1) ?Give an exact number.
Given Equation Differentiation: We are given the equation y4−2x=5. To find the second derivative of y with respect to x, we first need to differentiate both sides of the equation with respect to x. Differentiate y4−2x=5 with respect to x: dxd(y4)−dxd(2x)=dxd(5) 4y3⋅dxdy−2=0
First Derivative Calculation: Now, we need to solve for dxdy, which is the first derivative of y with respect to x.4y3⋅dxdy=2dxdy=4y32dxdy=2y31
Second Derivative Calculation: Next, we differentiate the first derivative with respect to x to find the second derivative.dxd(dxdy)=dxd(2y31)Using the chain rule and the quotient rule, we get:dx2d2y=2y3−6y2⋅(dxdy)2
Substitution for Given Point: We substitute the value of dxdy from the previous step into the expression for the second derivative.dx2d2y=−6y2×(2y31)2/(2y3)dx2d2y=−6y2×(4y61)/(2y3)dx2d2y=−6y2/(8y9)dx2d2y=−4y73
Final Result: Finally, we substitute the given point (−2,1) into the expression for the second derivative.(d2y)/(dx2) at the point (−2,1) is:(d2y)/(dx2)=−3/(4×17)(d2y)/(dx2)=−3/4
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