NIMCET Differentiation Previous Year Questions (PYQs) – Page 2 of 2

Let $y = [\cos(x^2)]^2$. Using the chain rule: 

 \[ \frac{dy}{dx} = 2\cos(x^2)\cdot \frac{d}{dx}\big(\cos(x^2)\big) \]  \[      = 2\cos(x^2)\cdot\big(-\sin(x^2)\cdot 2x\big) \] \[= -4x\,\cos(x^2)\sin(x^2). \] Using $\;2\sin u\cos u=\sin(2u)$ with $u=x^2$: \[ \frac{dy}{dx} = -2x\,\sin\!\big(2x^2\big). \]

Answer: ${\dfrac{dy}{dx}=-4x\,\sin\!\big(x^2\big) \cos (x^2)}$

$\dfrac{d}{dx}(x^3) = 3x^2$

$\dfrac{d}{dx}(e^x) = e^x$

$\dfrac{d}{dx}(3x) = 3$

$\dfrac{d}{dx}(\cot x) = -cosec^2 x$

Therefore, total derivative $= 3x^2 + e^x + 3 - cosec^2 x$

Answer: $\boxed{3x^2 + e^x + 3 - cosec^2 x}$ ✅

Let $y=-\log(\log x)$

$\dfrac{dy}{dx}=-\dfrac{d}{dx}[\log(\log x)]$

$\dfrac{d}{dx}[\log(\log x)]=\dfrac{1}{\log x}\cdot\dfrac{1}{x}$

Therefore, $\dfrac{dy}{dx}=-\dfrac{1}{x\log x}$

Answer: $\boxed{-\dfrac{1}{x\log x}},\; x>1$ ✅