Understanding the Derivative of Y x^2 √(7x - 14)^(1/3) / (1 - x^3^4)

In this detailed guide, we will explore the process of finding the derivative of the complex function Y x^2 √(7x - 14)^(1/3) / (1 - x^3^4). We will apply the quotient rule and the chain rule to achieve this, breaking down each step for clarity.

1. Defining the Function

Let's start by defining our function more formally:

Y frac{x^2 sqrt{7x - 14}^{1/3}}{1 - x^3^4}

2. Applying the Quotient Rule

The quotient rule states that if you have a function in the form Y frac{u}{v}, then the derivative Y' is given by:

Y' frac{uv' - vu'}{v^2}

Where u x^2 sqrt{7x - 14}^{1/3} and v 1 - x^3^4.

3. Breaking Down u and v

3.1. Finding u

u x^2 sqrt{7x - 14}^{1/3}

We will use the product rule to differentiate u: u x^2 sqrt{7x - 14}^{1/3} First, differentiate x^2: frac{du_1}{dx} 2x Second, differentiate sqrt{7x - 14}^{1/3} using the chain rule: Lets z sqrt{7x - 14} (7x - 14)^{1/2} z^{1/3} (7x - 14)^{1/6} First, differentiate z: frac{dz}{dx} frac{1}{2}(7x - 14)^{-1/2} cdot 7 frac{7}{2sqrt{7x - 14}} Now, using the chain rule: sqrt{7x - 14}^{1/3} frac{1}{3} frac{7x - 14}^{1/6} cdot frac{7}{2sqrt{7x - 14}} frac{7}{6} (7x - 14)^{-5/6}

Now, substituting back, we get:

u 2x sqrt{7x - 14}^{1/3} - x^2 cdot frac{7}{6} frac{1}{7x - 14}^{5/6}

3.2. Finding v

v 1 - x^3^4

We will differentiate v using the chain rule:

v 4 cdot (1 - x^3^3) cdot 3x^2 12x^2 (1 - x^3^3)

4. Applying the Quotient Rule Formula

Substituting u, u', v, v' into the quotient rule formula:

Y' frac{(2x sqrt{7x - 14}^{1/3} - x^2 cdot frac{7}{6} frac{1}{7x - 14}^{5/6})(1 - 12x^2 (1 - x^3^3))}{(1 - x^3^4)^2}

This expression can be further simplified, but it captures the derivative of the function. The first derivative of the function is:

Y' frac{(2x sqrt{7x - 14}^{1/3} - frac{7x^2}{6} (7x - 14)^{-5/6})(1 - 12x^2 (1 - x^3^3))}{(1 - x^3^4)^2}

5. Simplification and Conclusion

This is the first derivative. If you require the second derivative, we would need to differentiate Y' again, which would involve more complex calculations. Let me know if you want to proceed with that!

Related Keywords

derivative, quotient rule, chain rule