Find the inverse of the function y = (square root x) + 4x , then solve for its 1st order derivative.
3 answers
Find the inverse of the function y = (square root x) + 4x , then solve for its 1st order derivative.
switch variables and solve for y, and you get (after squaring)
16y^2 - (8x+1)y + x^2 = 0
y = [(8x+1)±√(16x+1)]/32
Pick the branch you want and take the derivative. Using the + branch,
y' = 1/4 (1 + 1/√(16x+1))
16y^2 - (8x+1)y + x^2 = 0
y = [(8x+1)±√(16x+1)]/32
Pick the branch you want and take the derivative. Using the + branch,
y' = 1/4 (1 + 1/√(16x+1))
or, using implicit,
16x^2 - 8xy - x + y^2 = 0
32x - 8y - 8xy' + 2yy' = 0
(2y-8x)y' = 8y-32x
dy/dx = (4-16x)/(y-8x)
dx/dy = (√x+4x-8x)/(4-16x)
= (1/4)(√x-4x)/(1-4x)
Check my math and see whether these agree.
16x^2 - 8xy - x + y^2 = 0
32x - 8y - 8xy' + 2yy' = 0
(2y-8x)y' = 8y-32x
dy/dx = (4-16x)/(y-8x)
dx/dy = (√x+4x-8x)/(4-16x)
= (1/4)(√x-4x)/(1-4x)
Check my math and see whether these agree.
1 answer