How do you differentiate $y = \frac{5 x^{2}}{4 x - 1}$ $y = (5x^2) / (4x-1)$ ?

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How do you differentiate $y = \frac{5 x^{2}}{4 x - 1}$ $y = (5x^2) / (4x-1)$ ?

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Answer 1 · 2016-06-09T08:01:17

$\frac{d y}{d x} = \frac{10 x (2 x - 1)}{{(4 x - 1)}^{2}}$ $dy/dx = (10x(2x-1))/ (4x-1)^2$

Just use the quotient rule and simplify.

Explanation:

$\frac{d y}{d x} = \frac{(4 x - 1) \cdot \frac{d}{d x} [5 x^{2}] - 5 x^{2} \cdot \frac{d}{d x} [4 x - 1]}{{(4 x - 1)}^{2}}$ $dy/dx = ((4x-1)* d/dx [ 5x^2 ] - 5x^2 * d/dx [ 4x-1 ] )/ (4x-1)^2$

$\frac{d y}{d x} = \frac{(4 x - 1) \cdot 10 x - 5 x^{2} \cdot 4}{16 x^{2} + 1 - 8 x}$ $dy/dx = ((4x-1)* 10x - 5x^2 * 4 )/ (16x^2+1-8x)$

$\frac{d y}{d x} = \frac{(40 x^{2} - 10 x) - 20 x^{2}}{16 x^{2} + 1 - 8 x}$ $dy/dx = ((40x^2-10x) - 20x^2 )/ (16x^2+1-8x)$

$\frac{d y}{d x} = \frac{10 x (2 x - 1)}{{(4 x - 1)}^{2}}$ $dy/dx = (10x(2x-1))/ (4x-1)^2$

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How do you differentiate $y = \frac{5 x^{2}}{4 x - 1}$ $y = (5x^2) / (4x-1)$ ?

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How do you differentiate $y = \frac{5 x^{2}}{4 x - 1}$ $y = (5x^2) / (4x-1)$ ?