Can someone differentiate this (calculus)?

For the function f(x) = g(x)h(x),

f '(x) = g(x) h '(x) + g '(x) h(x)

f '(x) = (5x - 5)^2 [(4 - x^3)^4]' + [(5x - 5)^2] (4 - x^3)^4

To find the derivative of (4 - x^3)^4, use the chain rule. Take the derivative of the outer function and multiply it by the derivative of the inner function.

[(4 - x^3)^4]' = 4(4 - x^3)^3 (-3x^2)

Same thing with (5x - 5)^2

[(5x - 5)^2]' = 2(5x - 5)(5)

Your final answer is:

f '(x) = (5x - 5)^2 (-12x^2) (4 - x^3)^3 + 10(5x - 5)(4 - x^3)^4

It would help organize if you split this into two 'parts' f(x) and g(x).

Set f(x) equal to (5x - 5)².

Set g(x) equal to (4 - x^3)^4.

Now you have f(x)g(x).

Using product rule, you get f'(x)g(x) + g'(x)f(x) (I know the prime signs are a little hard to see with this part).

Take the derivatives of f(x) and g(x) (to get f'(x) and g'(x)). Chain rule comes in here.

f'(x) = [(5x - 5)²]'

f'(x) = 2 * (5x - 5)^1 * (5x - 5)'

f'(x) = 2 * (5x - 5) * (5)

f'(x) = 50x - 50

g'(x) = [(4 - x^3)^4]'

g'(x) = 4 * (4 - x^3)^3 * (4 - x^3)'

g'(x) = 4 * (4 - x^3)^3 * (-3x²)

g'(x) = -12x² * (4 - x^3)^3

Substitute f(x), g(x), f'(x), and g'(x) into the product ruled function. You should get the following:

[50x - 50] * [(4 - x^3)^4] + [-12x² * (4 - x^3)^3] * [(5x - 5)²]

Now it's just expansion and combining like terms (yay algebra!). I know my calculus teacher would accept this form.

Cheers for my TI-89 Titanium :D

350x^13 - 650x^12 + 300x^11 - 4400x^10 + 8000x^9 - 3600x^8 + 19200x^7 - 33600x^6 + 14400x^5 - 32000x^4 + 51200x^3 - 19200x² + 12800x - 12800

You are right, you must use both the chain rule and the product rule simultaneously:

(5x-5)^2(4-x^3)^4

So first you want to set it up like any other product rule:

[(5x - 5)^2]' [(4 - x^3)^4] + [(5x - 5)^2] [(4 - x^3)^4]'

now differentiate where indicated above but use the chain rule:

[2(5x - 5) * 5] [(4 - x^3)^4] + [(5x - 5)^2] [4(4 - x^3)^3 * -3x^2]

now bring all you numbers and constants to front:

10(5x - 5) (4 - x^3)^4 - 12x^2[(5x - 5)^2] [(4 - x^3)^3]

y = (5x - 5)^2 * (4 - x^3)^4

Let u = (5x - 5)^2 and v = (4 - x^3)^4

Then, y = uv, and use the product rule : dy/dx = u(dv/dx) + v(du/dx).

So first time round, after substituting, we have :

dy/dx = (5x - 5)^2 * (d/dx)(4 - x^3)^4 + (4 - x^3)^4 * (d/dx)(5x - 5)^2

Now you have to calculate those two derivatives using the chain law.

In general, if w = [f(x)]^n, then dw/dx = n * [f(x)]^(n - 1) * f '(x)

(1) We already have : u = (5x - 5)^2

Substituting in the general equation, w = u, f(x) = 5x - 5 (hence, f '(x) = 5) and n = 2,

so, du/dx = 2 * (5x - 5)^1 * 5 = 10(5x - 5) = 50(x - 1).

(2) With v = (4 - x^3)^4, we have : w = v, f(x) = 4 - x^3 (hence, f '(x) = -3x^2) and n = 4.

Therefore, dv/dx = 4 * (4 - x^3)^3 * (-3x^2) = -12x^2(4 - x^3)^3

Now substitute du/dx and dv/dx back into dy/dx above :

dy/dx = (5x - 5)^2 * [-12x^2(4 - x^3)^3] + (4 - x^3)^4 * [50(x - 1)]

Simplify : dy/dx = 50(x - 1)(4 - x^3)^4 - 300x^2(x - 1)^2(4 - x^3)^3

Take out the factor, 50(x - 1)(4 - x^3)^3 :

dy/dx = 50(x - 1)(4 - x^3)^3[4 - x^3 - 6x^2(x - 1)]

= 50(x - 1)(4 - x^3)^3(-7x^3 + 6x^2 + 4)

In Calculus use d/dx and using Both Diffentional and integral with values you can solve it!

(5x-5)^2*4(4-x^3)^3(-3x^2)+(4-x^3)^4*2(5x-5)(5)

=-12x^2(5x-5)^2(4-x^3)^3+(50x-50)(4-3x)^4