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Exercises · Q12

Q.Differentiate y=5ex+2ln⁡(3x)y = 5e^x + 2\ln(3x) with respect to xx.

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Given: y=5ex+2ln⁡(3x)y = 5e^x+2\ln(3x).

Step 1 — Differentiate 5ex5e^x: since the exponent is plain xx, ddx(ex)=ex\dfrac{d}{dx}(e^x)=e^x directly, so ddx(5ex)=5ex\dfrac{d}{dx}(5e^x)=5e^x.

Step 2 — Differentiate 2ln⁡(3x)2\ln(3x): using ddx(ln⁡g(x))=g′(x)g(x)\dfrac{d}{dx}(\ln g(x))=\dfrac{g'(x)}{g(x)} with g(x)=3xg(x)=3x, g′(x)=3g'(x)=3: ddx(ln⁡(3x))=33x=1x\dfrac{d}{dx}(\ln(3x)) = \dfrac{3}{3x}=\dfrac1x, so ddx(2ln⁡(3x))=2x\dfrac{d}{dx}(2\ln(3x)) = \dfrac2x.

Step 3 — Combine: dydx=5ex+2x\dfrac{dy}{dx} = 5e^x+\dfrac2x. …

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