Calculus problem confusion?

Question

I'm trying to understand how to do this one calculus problem but haven't been able to do it with the exponents. Any help appreciated. Thanks.

Find d/dx [sin^4 (x^3)]

Answer 1

You have to keep using the chain rule. Take the derivative of what's on the outside, which is the exponent: 4 [sin (x³) ] ³

Now multiply that by the derivative of what's immediately inside:
4 [sin (x³)] ³ cos(x³)

And finally multiply it by the derivative of what's inside of the last term you just made:
4 [sin (x³)]³ * cos(x³) * 3x²

Simplify:
(12x²) [sin³(x³)] cos(x³)

Answer 2

f(x) = sin^4(x^3)

To solve this, you need knowledge of the chain rule.
You're also dealing with three functions here: the 4th power, the sine, and the x^3. d/dx x^4 = 4x^3, d/dx sin(x) = cos(x), and
d/dx x^3 = 3x^2.

I'm going to rewrite the function slightly to make all the functions more obvious.

f(x) = [sin(x^3)]^4

Therefore, by the chain rule (twice),

f'(x) = 4[sin(x^3)]^3 (cos(x^3)) (3x^2)

Answer 3

ok so you're trying to find the derivative of [ sin^4 (x^3) ], which looks like this: [sin(x^3)]^4. so you do the chain rule twice -->
d/dx [sin(x^3)]^4 = 4[sin(x^3)]^3 * cos(x^3) * 3x^2

Answer 4

y = [ sin ^(4) (x³) ]

Let u = x³

du / dx = 3x²

y = sin^(4) u

dy/du = 4 sin³ u . cos u

dy / dx = (dy / du) . (du /dx)

dy / dx = 4 sin ³ u . cos u . 3x²

dy / dx = 4 sin³ (x³ ) . cos (x³) . 3 x²

dy / dx= 12x² . sin ³ (x ³) . cos (x³)

Answer 5

15/x(x - 1)(x - 2) = A/x + B/(x - 1) + C/(x - 2) 15 = A(x - 1)(x - 2) + Bx (x - 2) + Cx(x - 1) 15 = A(x^2 - 3x + 2) + B(x^2 - 2x) + C(x^2 - x) 15 = x^2 (A + B + C) + x(-3A - 2B - C) + 2A 2A = 15 ==> A = 15/2 -3A - 2B - C = 0 ==> - 2B - C = 45/2 -----------(1) A + B + C = 0 ==> B + C = - 15/2----------------(2) Add eqns (1) and (2) -B = 15 ==> B = - 15 substitute in (2) C = -15/2 + 15 ==< C = 15/2 A = 15/2, B = -15 and C = 15/2

Answer 6

d/dx [sin^4 (x^3)] =
[4sin^3(x^3)][cos(x^3)](3x^2) =
(12x^2)[4sin^3(x^3)][cos(x^3)]