Given that $$\frac{2x^{2} - x^{3}}{\sqrt{x}}$$ can be written in the form $2x^{p} - x^{q}$, (a) write down the value of $p$ and the value of $q$ - Edexcel - A-Level Maths Pure - Question 8 - 2009 - Paper 1

To find $\frac{dy}{dx}$ for

$y = 5x^{4} - 3 + \frac{2x^{2} - x^{3}}{\sqrt{x}},$

we first rewrite the expression:

$y = 5x^{4} - 3 + 2x^{1} - x^{\frac{3}{2}}$

Now, differentiating term by term:

1. For $5x^{4}$, we get: $\frac{d}{dx}(5x^{4}) = 20x^{3}.$

2. For $-3$, the derivative is: $\frac{d}{dx}(-3) = 0.$

3. For $2x^{1}$, we get: $\frac{d}{dx}(2x^{1}) = 2.$

4. For $-x^{\frac{3}{2}}$, we apply the power rule: $\frac{d}{dx}(-x^{\frac{3}{2}}) = -\frac{3}{2}x^{\frac{1}{2}}.$

Combine these results:

$\frac{dy}{dx} = 20x^{3} + 2 - \frac{3}{2}x^{\frac{1}{2}}.$

Thus the final simplified expression for $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = 20x^{3} + 2 - \frac{3}{2}\sqrt{x}.$