If y-4x2x3-3- then dydx-

We have, y=4x^2x^3+3 Let, u=4x^2 and v=x^3+3 dudx=8x and dvdx=3x^2 Using quotient rule: dydx=d( uv)dx= v dudx u dvdxv^2 =(x^3+3)(8x) (4x^2)(3x^2)(x^3+3)^2 = 8x ...

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Standard XII

Physics

Basic Differentiation Rule

If y=4x2x3+3,...

Question

If $y = \frac{4 x^{2}}{x^{3} + 3},$ then $\frac{d y}{d x} =$

\frac{- 4 x^{4} + 24 x}{(x^{3} + 3)}

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\frac{- 4 x^{4} + 24 x}{(x^{3} + 3)^{2}}

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\frac{4 x^{3} + 24 x}{(x^{3} + 3)^{2}}

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

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Solution

The correct option is B $\frac{- 4 x^{4} + 24 x}{(x^{3} + 3)^{2}}$
We have,
$y = \frac{4 x^{2}}{x^{3} + 3}$
Let,
$u = 4 x^{2}$ and $v = x^{3} + 3$
$\frac{d u}{d x} = 8 x$ and $\frac{d v}{d x} = 3 x^{2}$
Using quotient rule:
$\frac{d y}{d x} = \frac{d (\frac{u}{v})}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$
$= \frac{(x^{3} + 3) (8 x) - (4 x^{2}) (3 x^{2})}{(x^{3} + 3)^{2}}$
$= \frac{8 x^{4} + 24 x - 12 x^{4}}{(x^{3} + 3)^{2}}$
$= \frac{- 4 x^{4} + 24 x}{(x^{3} + 3)^{2}}$

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Basic Differentiation Rule

Standard XII Physics

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If y=4x2x3+3, then dydx=

The correct option is B −4x4+24x(x3+3)2We have, y=4x2x3+3 Let, u=4x2 and v=x3+3 dudx=8x and dvdx=3x2 Using quotient rule: dydx=d(uv)dx=vdudx−udvdxv2 =(x3+3)(8x)−(4x2)(3x2)(x3+3)2 =8x4+24x−12x4(x3+3)2 =−4x4+24x(x3+3)2

If $y = \frac{4 x^{2}}{x^{3} + 3},$ then $\frac{d y}{d x} =$