Match List-I with List-II | List-I | List-II | |---|---| | (A) Perpetuity | (I) A person deposits a fixed amount every year in his bank account to…

Given: $y = \log_e(\sec e^{x^2})$y=loge​(secex2)

Since $\log_e = \ln$loge​=ln and $\sec \theta = \frac{1}{\cos \theta}$secθ=cosθ1​:

$$y = \ln\left(\frac{1}{\cos e^{x^2}}\right)$$y=ln(cosex21​)

Using the property $\ln\left(\frac{1}{a}\right) = -\ln(a)$ln(a1​)=−ln(a):

$$y = -\ln(\cos e^{x^2})$$y=−ln(cosex2)

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Differentiating with respect to $x$x, where $u = \cos e^{x^2}$u=cosex2:

$$\frac{dy}{dx} = -\frac{1}{\cos e^{x^2}} \cdot \frac{d}{dx}(\cos e^{x^2})$$dxdy​=−cosex21​⋅dxd​(cosex2)

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For $\frac{d}{dx}(\cos e^{x^2})$dxd​(cosex2), let $v = e^{x^2}$v=ex2:

$$\frac{d}{dx}(\cos e^{x^2}) = -\sin e^{x^2} \cdot \frac{d}{dx}(e^{x^2})$$dxd​(cosex2)=−sinex2⋅dxd​(ex2)

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For $\frac{d}{dx}(e^{x^2})$dxd​(ex2), let $w = x^2$w=x2:

$$\frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot 2x$$dxd​(ex2)=ex2⋅2x

$$\frac{d}{dx}(e^{x^2}) = 2xe^{x^2}$$dxd​(ex2)=2xex2

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Combining the results:

$$\frac{d}{dx}(\cos e^{x^2}) = -\sin e^{x^2} \cdot 2xe^{x^2}$$dxd​(cosex2)=−sinex2⋅2xex2

$$\frac{dy}{dx} = -\frac{1}{\cos e^{x^2}} \cdot (-\sin e^{x^2} \cdot 2xe^{x^2})$$dxdy​=−cosex21​⋅(−sinex2⋅2xex2)

$$\frac{dy}{dx} = \frac{\sin e^{x^2}}{\cos e^{x^2}} \cdot 2xe^{x^2}$$dxdy​=cosex2sinex2​⋅2xex2

Using $\frac{\sin \theta}{\cos \theta} = \tan \theta$cosθsinθ​=tanθ:

$$\frac{dy}{dx} = 2xe^{x^2} \tan e^{x^2}$$dxdy​=2xex2tanex2