Can you help me differentiate this?

Question

y = 4sinx * ln(x+3)^3

i know i have to use product rule first then chain rule for the second expression.

dy/dx = 4cosx [am i right?] * 3ln(x+3)^2

ok...where do i from here?

ok....i really have no idea what you two are talking about :(

*sigh*

thx for your help anyway. i'll have to ask my math teacher on monday.

cheers

Answer 1

Yes, you were doing fine. I assume that your parenthesis for the ln expression is like ln [(x+3)^3]. Following the rule for differentiating a product of two functions

dy/dx = 4 d/dx(sin(x)) ln[(x+3)^3] + 4 sin(x) d/dx(ln[(x+3)^3])
= 4 cos(x) ln[(x+3)^3] + 4 sin(x) d/dx(ln[(x+3)^3])

Now,

d/dx(ln[(x+3)^3]) = d/dx(3 ln(x+3)) = 3/(x+3)

Plugging this value in:
dy/dx = 4 cos(x) ln[(x+3)^3] + 4 sin(x) 3/(x+3) so

dy/dx = 4 cos(x) ln[(x+3)^3] +12 sin(x)/(x+3)

Answer 2

You need to use the product rule of calculus, as well as the chain rule at the same time.

Let f(x) = 4*sin(x) and let g(x) = [ln(x+3)]^3. Then,

y = f(x)*g(x). So just apply the product rule:

y' = f'(x)*g(x) + f(x)*g'(x).

f'(x) = 4*cos(x)

To find g'(x), you need to use the chain rule:

g'(x) = 3*[ln(x+3)]^2 * (1/(x+3))

(here, you use the power rule first, and then multiply by the derivative of the "inside", as in the chain rule).

Now just multiply together:

y' = 4*cos(x) * [ln(x+3)]^3 + 4*sin(x) * 3*[ln(x+3)]^2 * (1/(x+3))

Answer 3

y= 4sinx*3ln(x+3) = 12 sinx*ln(x+3)
y' = 12sinx *1/(x+3) + 12cosxln(x+1)
y' =12[ sinx/(x+3) + cosxln(x+1)]

Answer 4

y = 4sinx * ln(x+3)^3
dy/dx = 4cosx * ln(x+3)^3 + 4sinx * 3/(x+3)
dydx = 4(cosx)(ln(x+3)^3) + 12sinx/(x+3)

Answer 5

dy/dx =
(4 cosx).[ ln(x + 3)³ ] + 3.ln(x + 3)².(4 sinx)
4.ln (x + 3)² [ cosx.ln (x + 3) + 3 sin x ]