CUET UG 2025 — Mathematics Arithmetic

Given $y = 3^x + e^x + x^x + x^3$y=3x+ex+xx+x3

Finding $\frac{dy}{dx}$dxdy​ by differentiating each term:

$$\frac{d}{dx}(3^x) = 3^x \log_e 3$$dxd​(3x)=3xloge​3

$$\frac{d}{dx}(e^x) = e^x$$dxd​(ex)=ex

$$\frac{d}{dx}(x^x) = x^x(\log_e x + 1)$$dxd​(xx)=xx(loge​x+1)

$$\frac{d}{dx}(x^3) = 3x^2$$dxd​(x3)=3x2

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Combining all derivatives:

$$\frac{dy}{dx} = 3^x \log_e 3 + e^x + x^x(\log_e x + 1) + 3x^2$$dxdy​=3xloge​3+ex+xx(loge​x+1)+3x2

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Substituting $x = 3$x=3:

$$\frac{dy}{dx}\bigg|_{x=3} = 3^3 \log_e 3 + e^3 + 3^3(\log_e 3 + 1) + 3(3)^2$$dxdy​​x=3​=33loge​3+e3+33(loge​3+1)+3(3)2

$$= 27\log_e 3 + e^3 + 27(\log_e 3 + 1) + 27$$=27loge​3+e3+27(loge​3+1)+27

$$= 27\log_e 3 + e^3 + 27\log_e 3 + 27 + 27$$=27loge​3+e3+27loge​3+27+27

$$= e^3 + 54\log_e 3 + 54$$=e3+54loge​3+54

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Therefore, the value of $\frac{dy}{dx}$dxdy​ at $x = 3$x=3 is:

$$e^3 + 54\log_e 3 + 54$$e3+54loge​3+54