If y-x2-2x-3x-4- then dy dx-

Hereu(x)=x^2+2x,v(x)=3x 4 Using quotient rule, dy/dx= (3x 4)d(x^2+2x)/dx (x^2+2x)d(3x 4)/dx/(3x 4)^2 =(3x 4)(2x+2) (x^2+2x)3/(3x 4)^2 = 3x^2 8x 8/(3x 4)^2 ...

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Byju's Answer

Standard XIII

Physics

Basic Differentiation Rule

If y=x2+2x/3x...

Question

$I f y = \frac{(x^{2} + 2 x)}{(3 x - 4)}, t h e n \frac{d y}{d x} = ?$

$\frac{- 3 x^{2} + 4 x + 8}{(3 x - 4)^{2}}$

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$\frac{3 x^{2} - 8 x - 8}{(3 x - 4)^{2}}$

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$\frac{3 x^{2} + 8 x + 8}{(3 x - 4)^{2}}$

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none of these

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Solution

The correct option is B
$\frac{3 x^{2} - 8 x - 8}{(3 x - 4)^{2}}$

Here $u (x) = x^{2} + 2 x, v (x) = 3 x - 4$

$U s i n g q u o t i e n t r u l e,$

$\frac{d y}{d x} = \frac{(3 x - 4) \frac{d (x^{2} + 2 x)}{d x} - (x^{2} + 2 x) \frac{d (3 x - 4)}{d x}}{(3 x - 4)^{2}}$

$= \frac{(3 x - 4) (2 x + 2) - (x^{2} + 2 x) 3}{(3 x - 4)^{2}} = \frac{3 x^{2} - 8 x - 8}{(3 x - 4)^{2}}$

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If $y = x^{2} + 5 x^{\frac{3}{2}} + \frac{2}{x}, t h e n \frac{d y}{d x} =$ ?

$I f y = \frac{(x^{2} + 2 x)}{(3 x - 4)}, t h e n \frac{d y}{d x} = ?$

Q. If

y = \frac{(x^{2} + 2 x)}{(3 x - 4)}

, then find

\frac{d y}{d x}

If $y = [\frac{x^{2} + 2 x}{3 x - 4}], t h e n \frac{d y}{d x}$ =?

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If y=(x2+2x)(3x−4), then dydx=?

The correct option is B 3x2−8x−8(3x−4)2 Here u(x)=x2+2x,v(x)=3x−4 Using quotient rule, dydx=(3x−4)d(x2+2x)dx−(x2+2x)d(3x−4)dx(3x−4)2 =(3x−4)(2x+2)−(x2+2x)3(3x−4)2=3x2−8x−8(3x−4)2

$I f y = \frac{(x^{2} + 2 x)}{(3 x - 4)}, t h e n \frac{d y}{d x} = ?$