Question

Differentiate: y=√6x³+4x²​

Answer 1

Answer:

4x^2 (x - 1) - 6x^3 (x^2 - 1)

= 4x^2 (x - 1) - 6x^3 (x + 1)(x - 1)

= (x - 1)(4x^2 - 6x^3 (x + 1))

= -2x^2 (x - 1)(-2 + 3x(x + 1))

= 2x^2 (x - 1)(3x^2 + 3x - 2)

Solve for the roots of 3x^2 + 3x - 2 = 0

Using the quadratic formula,

x = ((-3) +/- sqrt (9 - 4*3 (-2)))/(2 * 3)

x = ((-3) +/- sqrt (33))/6

3x^2 + 3x - 2 = (x - (-3 + sqrt 33)/6)(x -(-3 - sqrt 33)/6)

= (x + (3 - sqrt 33)/6)(x - (3 + sqrt 33)/6)

2x^2 (x - 1)(3x^2 + 3x - 2)

= 2x^2 (x - 1)(x + (3 - sqrt 33)/6)(x - (3 + sqrt 33)/6)

Hope it was helpful

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Answer 2

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

$\green \odot \: \: \frac{d \: {x}^{n} }{dx} = n {x}^{n - 1} \\ \green \odot \: \: \: \frac{d \: \: \sqrt{x} }{dx} = \frac{1}{2 \sqrt{x} } \\ \\ now\\ \: \: \: \: \: \: \: y = \sqrt{6 {x}^{3} } + 4 {x}^{2} \\ \implies \: \: \frac{dy}{dx} = \frac{d \: ( \sqrt{6 {x}^{3} } + 4 {x}^{2}) }{dx} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{d \: \sqrt{6 {x}^{3} } }{dx} + \frac{d \: 4 {x}^{2} }{dx} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2 \sqrt{6 {x}^{3} } } \times \frac{d \: 6 {x}^{3} }{dx} + 4 \times 2 \times x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2 \sqrt{6 {x}^{3} } } \times 6 \times 3 \times {x}^{2} + 8x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{9 {x}^{2} }{ \sqrt{6 {x}^{3} } } + 8x$