Phrases Previous Year Questions (PYQs) – Differentiation – Page 1 of 2

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

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

Answer: ${\displaystyle \frac{dy}{dx}}=-4x\sin (x^2)\cos (x^2)$

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

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

${\displaystyle \frac{d}{dx}}(3x)=3$

${\displaystyle \frac{d}{dx}}(\cot x)=-cose{c}^{2}x$

Therefore, total derivative $=3x^2+e^x+3-cose{c}^{2}x$

Answer: $\boxed{{\displaystyle 3x^2+e^x+3-cose{c}^{2}x}}$ ✅

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

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

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

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

Answer: $\boxed{{\displaystyle -{\displaystyle \frac{1}{x\log x}}}},x>1$ ✅