Question

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!

Answer 1

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

Answer 2

Thanks so much. Could you help me with 2 & 3?

Answer 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'