sec^2(4x+y) * (4+y') = 4
4+y' = 4cos^2(4x+4y)
y' = 4cos^2(4x+4y)-4 = -4sin^2(4x+4y)
Find dy/dx by implicit differentiation given that
tan(4x+y)=4x
4 answers
from the ans choices i choose:
dy/dx= - 4x^2/x^2+1
is that right
dy/dx= - 4x^2/x^2+1
is that right
Looks good to me.
sec^2(4x+y) * (4+y') = 4
4+y' = 4cos^2(4x+4)
y' = 4cos^2(4x+y)-4 = -4sin^2(4x+y)
Then sin^2(4x+y)=(4x/sqrt(16x^2+1))^2
=16x^2/(16x^2+1)
So y' = -4(16x^2/(16x^2+1))
=-64x^2/(16x^2+1))
4+y' = 4cos^2(4x+4)
y' = 4cos^2(4x+y)-4 = -4sin^2(4x+y)
Then sin^2(4x+y)=(4x/sqrt(16x^2+1))^2
=16x^2/(16x^2+1)
So y' = -4(16x^2/(16x^2+1))
=-64x^2/(16x^2+1))