Find the derivative of y x - x 3- x -12 with respect to X-A- 2 x2-x-12-2 x3-x-13B- 3 x2-x-12-3 x3-x-13C- 3 x2-x-12-2 x3-x-13D- 3 x2-x-12-2 x3-x-13

We can rewrite this function as nbsp;y(x)=x^3 (x+1)^ 2 and apply product rule, dy/dx =(x+1)^ 2d/dx (x^3 )+x^3d/dx (x+1)^ 2 dy/dx =(x+1)^ 2.3x^2+x^3 ( 2) (x+1) ...

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Byju's Answer

Standard XI

Mathematics

Derivative from First Principle

Find the deri...

Question

Find the derivative of $y (x) = \frac{x^{3}}{(x + 1)^{2}}$ with respect to $x$ .

\frac{2 x^{2}}{(x + 1)^{2}} - \frac{2 x^{3}}{(x + 1)^{3}}

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\frac{3 x^{2}}{(x + 1)^{2}} - \frac{2 x^{3}}{(x + 1)^{3}}

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\frac{3 x^{2}}{(x + 1)^{2}} + \frac{2 x^{3}}{(x + 1)^{3}}

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\frac{3 x^{2}}{(x + 1)^{2}} - \frac{3 x^{3}}{(x + 1)^{3}}

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Solution

The correct option is B $\frac{3 x^{2}}{(x + 1)^{2}} - \frac{2 x^{3}}{(x + 1)^{3}}$
We can rewrite this function as $y (x) = x^{3} (x + 1)^{- 2}$ and apply product rule,
$\frac{d y}{d x} = (x + 1)^{- 2} \frac{d}{d x} (x^{3}) + x^{3} \frac{d}{d x} (x + 1)^{- 2} \frac{d y}{d x} = (x + 1)^{- 2} .3 x^{2} + x^{3} (- 2) (x + 1)^{- 3} \frac{d y}{d x} = \frac{3 x^{2}}{(x + 1)^{2}} - \frac{2 x^{3}}{(x + 1)^{3}}$

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Similar questions

Q. Solve the equations:

\frac{2 x^{3} - 3 x^{2} + x + 1}{2 x^{3} - 3 x^{2} - x - 1} = \frac{3 x^{3} - x^{2} + 5 x - 13}{3 x^{3} - x^{2} - 5 x + 13}

Q. The derivative of $(x^2+ln6)(\sqrt{x})$ will be

lim x \to 2 \frac{x^{3} + 3 x^{2} - 9 x - 2}{x^{3} - x - 6}

Q. Assertion (A): The function

f (x) = 2 x^{3} - 3 x^{2_{-}} 12 x + 8

has minimum value -12 at x = 2
Reason (R): For the fucntion

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 8, f^{'} (2) = 0

and

f^{''} (2) > 0

Find the Integral of the given function w.r.t x

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

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Find the derivative of y(x)=x3(x+1)2 with respect to x.

The correct option is B 3x2(x+1)2−2x3(x+1)3We can rewrite this function as y(x)=x3(x+1)−2 and apply product rule, dydx=(x+1)−2ddx(x3)+x3ddx(x+1)−2dydx=(x+1)−2.3x2+x3(−2)(x+1)−3dydx=3x2(x+1)2−2x3(x+1)3

Find the derivative of $y (x) = \frac{x^{3}}{(x + 1)^{2}}$ with respect to $x$ .