Using the differentiation rules and the derivative table, find the derivative of the function y=\sqrt{2^x+\sqrt{x}}

y' = \frac{1}{2\sqrt{2^x + \sqrt{x}}} \left( 2^x \ln(2) + \frac{1}{2 \sqrt{x}} \right)

We can use the Chain Rule.

  • Identify the outer function and the inner function
    We can rewrite the function as:


    y = (g(x))^{1/2}


    namely the composition f\circ g, where g(x) = 2^x + \sqrt{x} and the outer function is f(\cdot)=(\cdot)^\frac12

  • Apply the Chain Rule
    The derivative of y in x is


    y' = \frac{1}{2} (g(x))^{-1/2} \cdot g'(x)



  • Calculate the derivative of g g(x) = 2^x + \sqrt{x}
    The derivative of g(x) is the sum of the derivatives of the individual components (for linearity of the derivative)


    g'(x) = \frac{d}{dx} 2^x + \frac{d}{dx} \sqrt{x}


    where the derivatives of the elementary functions are

    \frac{d}{dx} 2^x = 2^x \ln(2)


    and

    \frac{d}{dx} \sqrt{x} = \frac{1}{2} x^{-1/2} = \frac{1}{2 \sqrt{x}}


    Combining these two derivatives:

    g'(x) = 2^x \ln(2) + \frac{1}{2 \sqrt{x}}

  • Replace g and g' in the Chain Rule

    y' = \frac{1}{2} (2^x + \sqrt{x})^{-1/2} \cdot \left( 2^x \ln(2) + \frac{1}{2 \sqrt{x}} \right)


    This is the final result.