Math Problem Statement

Jika y = √(2x + 6), maka y' = ...

Solution

Let's solve the problem in the image. The given equation is:

y=2x+6y = \sqrt{2x + 6}

We are tasked with finding the derivative yy'.

Step-by-step process:

  1. Rewrite the equation: y=(2x+6)1/2y = (2x + 6)^{1/2}

  2. Use the chain rule to differentiate: y=12(2x+6)1/2ddx(2x+6)y' = \frac{1}{2} (2x + 6)^{-1/2} \cdot \frac{d}{dx}(2x + 6)

  3. Differentiate 2x+62x + 6: ddx(2x+6)=2\frac{d}{dx}(2x + 6) = 2

  4. Combine the results: y=12(2x+6)1/22y' = \frac{1}{2} (2x + 6)^{-1/2} \cdot 2 Simplifying further: y=22(2x+6)1/2=(2x+6)1/2y' = \frac{2}{2} (2x + 6)^{-1/2} = (2x + 6)^{-1/2}

  5. Return the derivative to radical form: y=12x+6y' = \frac{1}{\sqrt{2x + 6}}

Conclusion:

The correct answer is option a, which matches the derivative we calculated:

12x+6\boxed{ \frac{1}{\sqrt{2x + 6}} }


Would you like more details or have any questions? Here are some related problems you could explore:

  1. How do you apply the chain rule to more complex functions?
  2. What happens if the function inside the square root is more complicated?
  3. How do you differentiate higher powers of functions, like (2x+6)n(2x + 6)^n?
  4. Can you use the product rule and chain rule together?
  5. What is the importance of simplifying derivatives after applying the chain rule?

Tip: When differentiating functions involving roots, rewriting them as fractional exponents makes it easier to apply the chain rule!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Algebra

Formulas

y = (2x + 6)^{1/2}
dy/dx = 1/2 * (2x + 6)^{-1/2} * d(2x+6)/dx

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12