NIMCET 2027 Differentiation Previous Year Questions (PYQs) – Page 1 of 2
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Showing 17 questions
$$
\frac{d^{2}x}{dy^{2}} \; = \; ?
$$
$
(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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Solution
If $y = f(x)$ is odd and differentiable on $(-\infty,\infty)$ such that
$f'(3) = -2$, then $f'(-3)$ equals:
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Solution
If $f(x)$ is odd, then:
$f(-x) = -f(x)$
Differentiate both sides:
$f'(-x)(-1) = -f'(x)$
$f'(-x) = f'(x)$
So derivative is even.
Therefore:
$f'(-3) = f'(3) = -2$
Find $ \displaystyle \frac{d}{dx}\left( \sqrt{x} - \frac{5}{\sqrt{x}} \right) $
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Solution
$ \displaystyle \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}},;;
\frac{d}{dx}\left(\frac{5}{\sqrt{x}}\right)= 5\cdot\left(-\frac{1}{2}x^{-3/2}\right) $
So
$ \displaystyle \frac{d}{dx}\left(\sqrt{x}-\frac{5}{\sqrt{x}}\right)
= \frac{1}{2\sqrt{x}} + \frac{5}{2}x^{-3/2} $
If $ f(x)=\displaystyle \int_{0}^{x} t\sin t, dt $, then $f'(x)$ is
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Solution
$ f'(x) = x\sin x $
Let $f(x)$ be a polynomial function of second degree and $f(1) = f(–1)$. If $a, b, c$ are in A.P. then $f'(a), f'(b), f'(c)$ are in.
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Solution
If x = a cos t, y = b sin t, then $\frac{d^{2}y}{dx^{2}}$ is
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Solution
The function $f(x)=\begin{cases}{{(1+2x)}^{1/x}} & {,x\ne0} \\ {{e}^2} & {,x=0}\end{cases}$, is
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Solution
NIMCET 2027
Online Test Series, Information About Examination,
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and More.
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NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
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