Question

(1 point) For the equation given below, evaluate ′ at the point (−1,2). (5−)^4+4^3=2433. ′ at (−1,2) =

Answer 1

Answer:

$\dfrac{343}{71}$

Step-by-step explanation:

Given the equation

$(5x-y)^4+4y^3=2433$

Find the derivative:

$((5x-y)^4+4y^3)'=(2433)'\\ \\4(5x-y)^3\cdot (5x-y)'+4\cdot 3y^2\cdot y'=0\\ \\4(5x-y)^3\cdot (5-y')+12y^2y'=0$

Substitute

$x=-1\\ \\y=2,$

then

$4(5\cdot (-1)-2)^3\cdot (5-y')+12\cdot 2^2\cdot y'=0\\ \\4(-5-2)^3(5-y')+48y'=0\\ \\4\cdot (-7)^3\cdot (5-y')+48y'=0\\ \\-1,372(5-y')+48y'=0\\ \\-6,860+1,372y'+48y'=0\\ \\1,420y'=6,860\\ \\y'=\dfrac{6,860}{1,420}=\dfrac{686}{142}=\dfrac{343}{71}$