𝑂𝑝𝑡𝑖𝑜𝑛(𝐶):0ex0exy=cos((1 x^2))^20ex0ex(1 x^2)^2=z0ex0exy=cos(z)0ex0ex Differentiate with respect to 'x'0ex0exdy/dx=d/dxcos(z)×dz/dz0ex0exdy/dx=d/dzcos(z)dz/dx0e ...

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Integration by Substitution

d/dx[cos1-x22...

Question

$(d / d x) [\cos {(1 - x 2)}^{2}] =$

$- 2 x (1 - x^{2}) \sin {(1 - x^{2})}^{2}$

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$- 4 x (1 - x^{2}) \sin {(1 - x^{2})}^{2}$

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$4 x (1 - x^{2}) \sin {(1 - x^{2})}^{2}$

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$n o n e o f t h e s e$

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Solution

The correct option is C
$4 x (1 - x^{2}) \sin {(1 - x^{2})}^{2}$

$O p t i o n (C) : y = \cos {((1 - x^{2}))}^{2} {(1 - x^{2})}^{2} = z y = \cos (z) D i f f e r e n t i a t e w i t h r e s p e c t t o' x' \frac{d y}{d x} = \frac{d}{d x} \cos (z) \times \frac{d z}{d z} \frac{d y}{d x} = \frac{d}{d z} \cos (z) \frac{d z}{d x} \frac{d y}{d x} = - \sin (z) - 4 x (1 - x^{2}) \frac{d y}{d x} = 4 x (1 - x^{2}) \sin {(1 - x^{2})}^{2} H e n c e o p t i o n (C) i s c o r r e c t$

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Similar questions

\int_{0}^{1} \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} d x

Q. Differentiate

\cos^{- 1} (\frac{x + \sqrt{1 - x^{2}}}{\sqrt{2}}), - 1 < x < 1

\int \frac{x}{4 + x^{4}} d x

is equal to

(a)

\frac{1}{4} \tan^{- 1} x^{2} + C

(b)

\frac{1}{4} \tan^{- 1} (\frac{x^{2}}{2})

(c)

\frac{1}{2} \tan^{- 1} (\frac{x^{2}}{2})

(d) none of these

\int \frac{d x}{(1 + x^{2})^{2}}

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

=

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(d/dx)[cos(1-x2)2]=

The correct option is C 4x(1-x2)sin(1-x2)2Option(C):y=cos1-x221-x22=zy=coszDifferentiatewithrespectto'x'dydx=ddxcosz×dzdzdydx=ddzcoszdzdxdydx=-sinz-4x1-x2dydx=4x1-x2sin1-x22Henceoption(C)iscorrect

$(d / d x) [\cos {(1 - x 2)}^{2}] =$