Math Problem Statement

Find the derivative of the function y = [x + (x + sin^2(x))^6]^7

Solution

The function provided is:

y=[x+(x+sin2(x))6]7y = \left[ x + (x + \sin^2(x))^6 \right]^7

To find the derivative dydx\frac{dy}{dx}, we'll need to use the chain rule and the product rule in the following steps.

Step 1: Apply the Chain Rule

The outermost part of the function is [u]7[u]^7, where u=x+(x+sin2(x))6u = x + (x + \sin^2(x))^6. The derivative of u7u^7 with respect to uu is:

ddx(u7)=7u6dudx\frac{d}{dx}\left( u^7 \right) = 7u^6 \cdot \frac{du}{dx}

Step 2: Find dudx\frac{du}{dx}

Now, we need to differentiate u=x+(x+sin2(x))6u = x + (x + \sin^2(x))^6. This will also require the chain rule. First, split uu into two parts:

  1. The derivative of xx with respect to xx is 1.
  2. The second term is (x+sin2(x))6(x + \sin^2(x))^6, so we need to apply the chain rule again.

Let v=x+sin2(x)v = x + \sin^2(x), so:

ddx(v6)=6v5dvdx\frac{d}{dx} (v^6) = 6v^5 \cdot \frac{dv}{dx}

Now, find dvdx\frac{dv}{dx}:

dvdx=ddx(x+sin2(x))=1+2sin(x)cos(x)=1+sin(2x)\frac{dv}{dx} = \frac{d}{dx} (x + \sin^2(x)) = 1 + 2\sin(x)\cos(x) = 1 + \sin(2x)

Thus,

dudx=1+6(x+sin2(x))5(1+sin(2x))\frac{du}{dx} = 1 + 6(x + \sin^2(x))^5 \cdot (1 + \sin(2x))

Step 3: Multiply and Simplify

Finally, we can substitute this expression for dudx\frac{du}{dx} into the chain rule result from step 1:

dydx=7[x+(x+sin2(x))6]6(1+6(x+sin2(x))5(1+sin(2x)))\frac{dy}{dx} = 7\left[ x + (x + \sin^2(x))^6 \right]^6 \cdot \left( 1 + 6(x + \sin^2(x))^5 \cdot (1 + \sin(2x)) \right)

This is the derivative of the given function.


Would you like to see a step-by-step solution of any particular part or need further clarification?

Related Questions:

  1. What are the key concepts in applying the chain rule?
  2. How do you differentiate functions with powers, like (xn)(x^n)?
  3. Can you explain how the sine and cosine functions are derived?
  4. What is the significance of the derivative in a real-world context?
  5. How would you compute the second derivative of this function?

Tip:

When solving complex derivatives, always break down the problem into smaller parts and apply the chain rule step-by-step.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Chain Rule
Trigonometric Functions

Formulas

Chain Rule: (f(g(x)))' = f'(g(x)) * g'(x)
Derivative of sin^2(x): d/dx[sin^2(x)] = 2sin(x)cos(x) = sin(2x)

Theorems

Chain Rule
Product Rule

Suitable Grade Level

Undergraduate Calculus (Grades 11-12 and above)