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nimcet Previous Year Questions (PYQs)
nimcet Differentiation PYQ
nimcet PYQ
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nimcet Mathematics PYQnimcet Differentiation (Disha JEE) PYQ
$$
\frac{d^{2}x}{dy^{2}} \; = \; ?
$$
$
(a) -\left(\dfrac{d^{2}y}{dx^{2}}\right)^{-1}\left(\dfrac{dy}{dx}\right)^{-3}
$
$ (b) \left(\dfrac{d^{2}y}{dx^{2}}\right)\left(\dfrac{dy}{dx}\right)^{-2}
$
$
(c) -\left(\dfrac{d^{2}y}{dx^{2}}\right)\left(\dfrac{dy}{dx}\right)^{-3}
$
$
(d) \left(\dfrac{d^{2}y}{dx^{2}}\right)^{-1}
$
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nimcet Mathematics PYQnimcet Differentiation (Disha JEE) PYQ
Solution
nimcet PYQ
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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ
The function $f(x)=\begin{cases}{{(1+2x)}^{1/x}} & {,x\ne0} \\ {{e}^2} & {,x=0}\end{cases}$, is
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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ
Solution
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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ
If $x^my^n$=$(x+y)^{m+n}$, then $\frac{dy}{dx}$ is
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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ
Solution
nimcet PYQ
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nimcet Previous Year PYQnimcet NIMCET 2021 PYQ
If $y={\tan }^{-1}\lgroup{\frac{3x-{x}^3}{1-3{x}^2}}\rgroup\, ,\, \frac{-1}{\sqrt[]{3}}{\lt}x{\lt}\frac{1}{\sqrt[]{3}}$ then $\frac{dy}{dx}$ is
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nimcet Previous Year PYQnimcet NIMCET 2021 PYQ
Solution
Incorrect questionnimcet PYQ
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nimcet Previous Year PYQnimcet NIMCET 2021 PYQ
If $y=\sin ^{-1}(\frac{{x}^2+1}{\sqrt[]{1+3{x}^2+{x}^4}}),\, (x>0),$ then $\frac{dy}{dx}$=
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nimcet Previous Year PYQnimcet NIMCET 2021 PYQ
Solution
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nimcet Previous Year PYQnimcet NIMCET 2018 PYQ
Differential coefficient of 
with respect to 
to
with respect to toGo to Discussion
nimcet Previous Year PYQnimcet NIMCET 2018 PYQ
Solution
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nimcet Previous Year PYQnimcet NIMCET 2017 PYQ
= 2 (cos x2) sin x2 .2x
If y = cos2 x2 , find dx/dy
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nimcet Previous Year PYQnimcet NIMCET 2017 PYQ
Solution
Let y = (cos x2)2 = 2 (cos x2) sin x2 .2x = -4 (cos x2) sin x2
nimcet PYQ
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nimcet Previous Year PYQnimcet NIMCET 2017 PYQ
= 3x2 + ex + 3x log3 - cosec2 x
The derivative of (x3 + ex + 3x + cotx) with respect to x is
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nimcet Previous Year PYQnimcet NIMCET 2017 PYQ
Solution
Let y=x3 + ex + 3x + cotxNow,
= 3x2 + ex + 3x log3 - cosec2 xnimcet PYQ
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nimcet Previous Year PYQnimcet NIMCET 2017 PYQ
= - 
(log (log x))
Differentiate [- log(log x), x > 1] with respect to x
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nimcet Previous Year PYQnimcet NIMCET 2017 PYQ
Solution
Let y = - log(log x), x>1On differentiating both sides,
= - (log (log x))= 
(log x)
(log x)= 
= 
nimcet PYQ
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nimcet Previous Year PYQnimcet NIMCET 2015 PYQ
With the usual notation $\frac{d^{2}x}{dy^{2}}$
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nimcet Previous Year PYQnimcet NIMCET 2015 PYQ
Solution
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