Math Problem Statement

Find the derivative using the chain rule

Solution

To find the derivative of the function y=x2sin4(x)+xcos2(x)y = x^2 \sin^4(x) + x \cos^{-2}(x) with respect to xx, we will use the chain rule and the product rule.

Step 1: Differentiate y=x2sin4(x)y = x^2 \sin^4(x)

We need to apply the product rule since this is a product of two functions: u(x)=x2u(x) = x^2 and v(x)=sin4(x)v(x) = \sin^4(x).

  • The derivative of u(x)=x2u(x) = x^2 is u(x)=2xu'(x) = 2x.
  • To differentiate v(x)=sin4(x)v(x) = \sin^4(x), apply the chain rule: ddxsin4(x)=4sin3(x)cos(x)\frac{d}{dx} \sin^4(x) = 4\sin^3(x) \cdot \cos(x) (since the derivative of sin(x)\sin(x) is cos(x)\cos(x)).

Now, apply the product rule: ddx(x2sin4(x))=u(x)v(x)+u(x)v(x)=2xsin4(x)+x24sin3(x)cos(x)\frac{d}{dx} \left( x^2 \sin^4(x) \right) = u'(x)v(x) + u(x)v'(x) = 2x \sin^4(x) + x^2 \cdot 4\sin^3(x) \cdot \cos(x) Simplifying: =2xsin4(x)+4x2sin3(x)cos(x)= 2x \sin^4(x) + 4x^2 \sin^3(x) \cos(x)

Step 2: Differentiate y=xcos2(x)y = x \cos^{-2}(x)

Again, we apply the product rule where u(x)=xu(x) = x and v(x)=cos2(x)v(x) = \cos^{-2}(x).

  • The derivative of u(x)=xu(x) = x is u(x)=1u'(x) = 1.
  • To differentiate v(x)=cos2(x)v(x) = \cos^{-2}(x), we can rewrite it as cos2(x)=sec2(x)\cos^{-2}(x) = \sec^2(x). ddxcos2(x)=ddxsec2(x)=2sec2(x)sec(x)tan(x)=2sec3(x)tan(x)\frac{d}{dx} \cos^{-2}(x) = \frac{d}{dx} \sec^2(x) = 2\sec^2(x) \cdot \sec(x) \cdot \tan(x) = 2\sec^3(x) \tan(x)

Now, apply the product rule: ddx(xcos2(x))=u(x)v(x)+u(x)v(x)=cos2(x)+x2sec3(x)tan(x)\frac{d}{dx} \left( x \cos^{-2}(x) \right) = u'(x)v(x) + u(x)v'(x) = \cos^{-2}(x) + x \cdot 2\sec^3(x) \tan(x) Simplifying: =cos2(x)+2xsec3(x)tan(x)= \cos^{-2}(x) + 2x \sec^3(x) \tan(x)

Final Step: Combine the derivatives

Now, add the derivatives from the two parts: dydx=2xsin4(x)+4x2sin3(x)cos(x)+cos2(x)+2xsec3(x)tan(x)\frac{dy}{dx} = 2x \sin^4(x) + 4x^2 \sin^3(x) \cos(x) + \cos^{-2}(x) + 2x \sec^3(x) \tan(x)

This is the derivative of the given function using the chain rule and product rule.

Would you like more details or have any questions?

Related Questions:

  1. How does the derivative behave as xx approaches 0?
  2. What are the critical points of this function?
  3. Can you simplify the derivative further in specific cases?
  4. How does the derivative affect the concavity of the function?
  5. What happens to the derivative when x=π2x = \frac{\pi}{2}?

Tip:

When applying the chain rule, always remember to multiply by the derivative of the inner function, especially when dealing with trigonometric identities.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Product Rule
Trigonometric Functions

Formulas

Product Rule
Chain Rule
Derivatives of Trigonometric Functions

Theorems

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Suitable Grade Level

Advanced High School / College