$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\times\frac{\text{dt}}{\text{dx}}=\frac{\text{g}'(\text{t})}{\text{f}'(\text{t})},$ where $\text{f}'(\text{t})\neq0.$
On the basis of above information, answer the following questions.
- The derivative of $\text{f}(\tan\text{x})\text{w.r.t.}\text{ g}(\sec\text{x})\text{ at}\text{ x}=\frac{\pi}{4},$ where f'(1) = 2 and $\text{g}'(\sqrt{2})=4,$ is:
- $\frac{1}{\sqrt{2}}$
- ${\sqrt{2}}$
- 1
- 0
- The derivative of $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$ is:
- -1
- 1
- 2
- 4
- The derivative of $\text{e}^{\text{x}^3}$ with respect to log x is:
- $\text{e}^{\text{x}^3}$
- $3\text{x}^22\text{e}^{\text{x}^3}$
- $3\text{x}^3\text{e}^{\text{x}^3}$
- $3\text{x}^2\text{e}^{\text{x}^3}+3\text{x}$
- The derivative of $\cos^{-1}(2\text{x}^2-1)\text{w.r.t.}\cos^{-1}\text{x}$ is:
- $2$
- $\frac{-1}{2\sqrt{1-\text{x}^2}}$
- $\frac{2}{\text{x}}$
- $1-\text{x}^2$
- If $\text{y}=\frac{1}{4}\mu^4$ and $\mu=\frac{2}{3}\text{x}^3+5,$ then $\frac{\text{dy}}{\text{dx}}=$
- $\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$
- $\frac{2}{7}\text{x}^2(2\text{x}^3+15)^3$
- $\frac{2}{27}\text{x}(2\text{x}^3+5)^3$
- $\frac{2}{7}(2\text{x}^3+15)^3$









Based on the above information, answer the following questions.