i keep on doing something wrong on this problem,i guess im forgetting to add or subtract some number.

use f'(x)=lim h-->0 f(x+h)-f(x)/h
to find the limit: lim h-->0

sin^(3)4(x+h)-sin^(3)4x/h

To find the limit of the given expression, we can follow these steps:

Step 1: Apply the limit definition of the derivative.
Start by writing down the derivative formula f'(x) = lim(h->0) [f(x+h) - f(x)] / h.

Step 2: Substitute the given function into the formula.
In this case, the function f(x) = sin^3(4x). So, we substitute this function into the formula: f'(x) = lim(h->0) [sin^3(4(x+h)) - sin^3(4x)] / h.

Step 3: Simplify the expression.
Expand sin^3(4(x+h)) and sin^3(4x):
f'(x) = lim(h->0) [(sin(4x+4h))^3 - (sin(4x))^3] / h.

Using the identity (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3, we can rewrite the numerator:
f'(x) = lim(h->0) [sin^3(4x) + 3sin^2(4x)(4h) + 3sin(4x)(4h)^2 + (4h)^3 - sin^3(4x)] / h.

Step 4: Simplify further.
Combine like terms in the numerator:
f'(x) = lim(h->0) [3sin^2(4x)(4h) + 3sin(4x)(4h)^2 + (4h)^3] / h.

Step 5: Cancel out h in the numerator and denominator.
Since h is common in each term of the numerator, we can cancel it out:
f'(x) = lim(h->0) 3sin^2(4x) + 3sin(4x)(4h) + (4h)^2.

Step 6: Take the limit as h goes to 0.
Now take the limit as h approaches 0:
f'(x) = 3sin^2(4x) + 0 + 0.

Step 7: Final answer.
The final answer is f'(x) = 3sin^2(4x).

To summarize, the limit of the expression (sin^3(4(x+h)) - sin^3(4x)) / h as h approaches 0 is equal to 3sin^2(4x).