CUET UG 2025 — Mathematics Calculus

The expression $e^{2\log_e x^3}$e2loge​x3 can be simplified before differentiating.

Using the logarithm power rule $\log_e x^3 = 3\log_e x$loge​x3=3loge​x:

$$2\log_e x^3 = 2 \times 3\log_e x$$2loge​x3=2×3loge​x

$$= 6\log_e x$$=6loge​x

The expression becomes:

$$e^{6\log_e x}$$e6loge​x

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Rewriting $6\log_e x$6loge​x as $\log_e x^6$loge​x6:

$$e^{6\log_e x} = e^{\log_e x^6}$$e6loge​x=eloge​x6

Using the property that $e^{\log_e a} = a$eloge​a=a (since $e$e and $\log_e$loge​ are inverse functions):

$$e^{\log_e x^6} = x^6$$eloge​x6=x6

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Applying the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$dxd​(xn)=nxn−1:

$$\frac{d}{dx}(x^6) = 6x^5$$dxd​(x6)=6x5

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Therefore, the answer is option 2: $6x^5$6x5