Ff 5x2 + y4 = 21 then evaluate the second derivative of y with respect to x when x = 2 and y = 1. round your answer to two decimal places. use the hyphen symbol, -, for negative values. answer asap

It is given that ,

Differentiating both sides with respect to x, we get

→4 y³ y' = 0 - 10 x , where y' is first derivative of y w.r.t  x.

→4 y³ y' =  - 10 x

When x=2, y=1, substituting these values in above equation

→4× 1³×y'= -10×2

→ y'= -20/4= -5

Differentiating again w.r.t  x,the above equation we get

→4 [ 3 y²y'² + y³ y''] = -10 where y'' is double derivative of y w.r.t x.

→3 y²y'² + y³ y'' = -10/4= -5/2

Substituting x=2, y=1, y'= -5,in above equation we get

→ 3 ×1²×(-5)² + 1³× y'' = -5/2

→  75 + y'' = -5/2

→ y''= -5/2 - 75

→ y'' =

→ y''=  -77.50(Answer)

g(x) = x² + 2

step-by-step explanation:

it is given that the graphs are the same shape.

the only difference is that the red graph is 2 units higher than the blue graph.

that is to say that the red graph is simply the blue graph that has been moved vertically by 2 units in the positive y direction.

hence

g(x) = f(x) + 2

or

g(x) = x² + 2

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2