find dy/dx
1. y=(lnx)^3
2. y=x^4lnx
3. y=sin^3(lnx)
1. use chain rule
u = lnx
y = u^3
dy/dx = 3u^2 du/dx = 3(lnx)^2/x
2. use product rule
f=x^4
g=lnx
y = fg
y' = f'g + fg' = 4x^3 lnx + x^4/x = x^3(4lnx + 1)
3. use chain rule
u = lnx
v = sin(u)
y = v^3
dy/dx = dy/dv dv/du du/dx
= 3v^2 * cos(u) * 1/x
= 3sin^2(lnx) * cos(lnx) * 1/x
To find dy/dx, we will need to use the chain rule of differentiation, which states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).
Let's find the derivatives for each of the given functions:
1. y = (lnx)^3
Here, we have y = f(g(x)), where f(u) = u^3 and g(x) = lnx.
Applying the chain rule, we have:
dy/dx = f'(g(x)) * g'(x)
f'(u) = 3u^2 and g'(x) = 1/x
Therefore,
dy/dx = (3(lnx)^2) * (1/x)
= 3(lnx)^2/x
2. y = x^4lnx
Again, we have y = f(g(x)), where f(u) = u * ln(u) and g(x) = x^4.
Applying the chain rule:
dy/dx = f'(g(x)) * g'(x)
f'(u) = ln(u) + 1 and g'(x) = 4x^3
Therefore,
dy/dx = (ln(x^4) + 1) * 4x^3
= 4x^3 (lnx + 4)
3. y = sin^3(lnx)
This time, we have y = f(g(x)), where f(u) = u^3 and g(x) = sin(lnx).
Applying the chain rule:
dy/dx = f'(g(x)) * g'(x)
f'(u) = 3u^2 and g'(x) = cos(lnx) * (1/x)
Therefore,
dy/dx = (3(sin(lnx))^2) * (cos(lnx) * (1/x))
= 3(sin(lnx))^2 * cos(lnx) / x
So, the derivatives are:
1. dy/dx = 3(lnx)^2/x
2. dy/dx = 4x^3 (lnx + 4)
3. dy/dx = 3(sin(lnx))^2 * cos(lnx) / x