a. y=ln(uv)
y'=1/(uv)*(uv'+vu')
ok, u=2x^3 u'=6x^2
so y'=1/(2x^3*2e^x) * (2x^3*2e^x + 6x^2*2e^x(
y'=(3+x)/x
Find the derivatives of
1. ln[(2x^3⋅2e^x)]
2. xy = cot(xy)
3. y = cos(x^3 + 27)
I don't really know where to start with these.
The answer is (3+x)/(x) for the first, y/-x for the second, and 35. -3x^2 sin(x^3 + 27) but I have no idea how they got there.
Thank you!
3 answers
Thanks so much. Could you help me with 2 & 3?
just do the same stuff:
xy = cot(xy)
y + xy' = -csc^2(xy) (y + xy')
y'(x + xcsc^2(xy)) = -ycsc^2(xy)-y
xy' (1+csc^2(xy)) = -y(1+csc^2(xy))
xy' = -y
y' = -y/x
The other is just a simple chain rule.
y = cos(u), so
y' = -sin(u) u'
xy = cot(xy)
y + xy' = -csc^2(xy) (y + xy')
y'(x + xcsc^2(xy)) = -ycsc^2(xy)-y
xy' (1+csc^2(xy)) = -y(1+csc^2(xy))
xy' = -y
y' = -y/x
The other is just a simple chain rule.
y = cos(u), so
y' = -sin(u) u'
1 answer