Question

y=2x square -3x + 1/underoot of x-3 ______differentiate it w.r.t x

Answer 1

Answer on Question #47296 – Math – Other

Question:

$y = 2x$ square $-3x + 1$/underoot of x-3 ______ differentiate it w.r.t x.

Answer:

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

Derivative:

$$\frac{d}{dx} \left(2x^2 - 3x + \frac{1}{\sqrt{x - 3}}\right) = 4x - \frac{1}{2(x - 3)^{3/2}} - 3$$ $$\begin{aligned} \frac{d}{dx} f(x) &= \frac{d}{dx} \left(2x^2 - 3x + \frac{1}{\sqrt{x - 3}}\right) \\ &= 2 \cdot \frac{d}{dx} \left(x^2\right) + \frac{d}{dx} \left(\frac{1}{\sqrt{x - 3}}\right) - 3 \\ &= \frac{-\frac{d}{dx} (\sqrt{x - 3})}{x - 3} + 2 \cdot 2x - 3 \\ &= 4x - \frac{\frac{1}{2 \cdot \sqrt{x - 3}} \cdot \frac{d}{dx} (x - 3)}{x - 3} - 3 \\ &= 4x - \frac{1}{2 \cdot (x - 3)^{\frac{3}{2}}} - 3 \end{aligned}$$

Answer:

$$\frac{d}{dx} \left(2x^2 - 3x + \frac{1}{\sqrt{x - 3}}\right) = 4x - \frac{1}{2(x - 3)^{3/2}} - 3$$

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