Question

If $\frac{dy}{dx} = \frac{1}{8\left(\sqrt{16+\sqrt{25+\sqrt{x}}}\right)\left(\sqrt{25+\sqrt{x}}\right)\sqrt{x}}$, then $y =$

• A. $\sqrt{16+\sqrt{25+\sqrt{x}}}+C$
• B. $\sqrt{16+\sqrt{25+\sqrt{x}}}+x+C$
• C. $\sqrt{16+\sqrt{25+\sqrt{x}}}+x^{2}+C$
• D. $x\sqrt{16+\sqrt{25+\sqrt{x}}}+C$
• E. $x^{2}\sqrt{16+\sqrt{25+\sqrt{x}}}+C$

Hint:

Look for nested square roots; the derivative of $\sqrt{f(x)}$ is $\frac{f'(x)}{2\sqrt{f(x)}}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given the derivative \( \frac{dy}{dx} \) and asked to find the original function y. This requires finding the antiderivative, or integral, of the given expression. However, the expression for the derivative is very complex, suggesting that direct integration would be difficult. An alternative approach is to differentiate the options and see which one matches the given \( \frac{dy}{dx} \).
Step 2: Key Formula or Approach:
We will use the chain rule for differentiation to find the derivative of the function given in option (A).
The chain rule: \( \frac{d}{dx}f(g(h(x))) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \).
The power rule for differentiation: \( \frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx} \).
Step 3: Detailed Explanation:
Let's test option (A). Let \( y = \sqrt{16+\sqrt{25+\sqrt{x}}} \).
We need to find \( \frac{dy}{dx} \). This is a nested chain rule problem.
Let's differentiate from the outside in:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{16+\sqrt{25+\sqrt{x}}}} \cdot \frac{d}{dx}\left(16+\sqrt{25+\sqrt{x}}\right) \] The derivative of the constant 16 is 0. So we continue with the inner part:
\[ = \frac{1}{2\sqrt{16+\sqrt{25+\sqrt{x}}}} \cdot \frac{1}{2\sqrt{25+\sqrt{x}}} \cdot \frac{d}{dx}(25+\sqrt{x}) \] The derivative of the constant 25 is 0. We differentiate the innermost part:
\[ = \frac{1}{2\sqrt{16+\sqrt{25+\sqrt{x}}}} \cdot \frac{1}{2\sqrt{25+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \] Now, multiply all the terms together:
\[ \frac{dy}{dx} = \frac{1}{2 \cdot 2 \cdot 2 \cdot \sqrt{x} \sqrt{25+\sqrt{x}} \sqrt{16+\sqrt{25+\sqrt{x}}}} \] \[ \frac{dy}{dx} = \frac{1}{8\sqrt{x}\sqrt{25+\sqrt{x}}\sqrt{16+\sqrt{25+\sqrt{x}}}} \] This result perfectly matches the given expression for \( \frac{dy}{dx} \). Therefore, the integral of the given expression is the function from option (A), plus an arbitrary constant C.
Step 4: Final Answer:
The function is \( y = \sqrt{16+\sqrt{25+\sqrt{x}}} + C \).