Find the derivative of the function

y=(6x+(9x+arcsin(4x))^4)^3

Answer: y′=
  




To find the derivative of \( y = \left(6x + \left(9x + \arcsin(4x)\right)^4\right)^3 \), we'll apply the chain rule and product rule step by step. Here's how it's done:

### Step 1: Apply the Chain Rule
Let \( u = 6x + \left(9x + \arcsin(4x)\right)^4 \). Then, \( y = u^3 \). 

Using the chain rule:
\[
y' = 3u^2 \cdot u'
\]

### Step 2: Find \( u' \)
Now, we need to differentiate \( u \):
\[
u = 6x + \left(9x + \arcsin(4x)\right)^4
\]

Let \( v = 9x + \arcsin(4x) \). Therefore:
\[
u = 6x + v^4
\]

The derivative of \( u \) with respect to \( x \) is:
\[
u' = 6 + 4v^3 \cdot v'
\]

### Step 3: Find \( v' \)
To find \( v' \), differentiate \( v = 9x + \arcsin(4x) \):
\[
v' = 9 + \frac{d}{dx}(\arcsin(4x))
\]

The derivative of \( \arcsin(4x) \) is:
\[
\frac{d}{dx}(\arcsin(4x)) = \frac{4}{\sqrt{1 - (4x)^2}} = \frac{4}{\sqrt{1 - 16x^2}}
\]

So:
\[
v' = 9 + \frac{4}{\sqrt{1 - 16x^2}}
\]

### Step 4: Substitute \( v' \) into \( u' \)
\[
u' = 6 + 4v^3 \left(9 + \frac{4}{\sqrt{1 - 16x^2}}\right)
\]

### Step 5: Substitute \( u' \) into \( y' \)
\[
y' = 3 \left(6x + \left(9x + \arcsin(4x)\right)^4\right)^2 \cdot \left(6 + 4\left(9x + \arcsin(4x)\right)^3 \cdot \left(9 + \frac{4}{\sqrt{1 - 16x^2}}\right)\right)
\]

This is the derivative of the function.

Would you like further details on any part of this process or have any specific questions?

### Related Questions:
1. How do you apply the chain rule for nested functions?
2. What is the derivative of the arcsine function?
3. How can you simplify the expression involving derivatives of trigonometric functions?
4. What is the product rule and when is it used?
5. How do you find derivatives of composite functions involving powers?

**Tip:** When working with derivatives of composite functions, always break them down into simpler components using substitution to systematically apply the chain rule.

Find the derivative of the function y=(6x+(9x+arcsin(4x))^4)^3

Find the derivative of the function y = (6x + (9x + arcsin(4x))^4)^3 step by step using the chain rule, product rule, and differentiation of trigonometric functions. This calculus problem involves composite functions and arcsin derivatives, making it suitable for high school seniors and early college students.

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