JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Analyzing +4 | -1

y = \left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right) \ldots \left( {1 + {x^{2n}}} \right),

then the value of

{\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}}

View answer and explanation

Correct option: C

Option C (1) is correct because after expanding the product and differentiating, the value of the derivative at x=0 is 1.

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f(x)=|x|^{|\tan x|}

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x y=1

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f(x)=[x] \sin (\pi x)

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y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

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v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}