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Exercises · Q9

Q.Differentiate f(x)=2x2+3xf(x) = 2x^2 + 3x from first principles.

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✓ Free question

Given: f(x)=2x2+3xf(x) = 2x^2+3x.

Step 1 — Write f(x+Δx)f(x+\Delta x): f(x+Δx)=2(x+Δx)2+3(x+Δx)=2x2+4xΔx+2(Δx)2+3x+3Δxf(x+\Delta x) = 2(x+\Delta x)^2 + 3(x+\Delta x) = 2x^2+4x\Delta x+2(\Delta x)^2+3x+3\Delta x.

Step 2 — Subtract f(x)=2x2+3xf(x) = 2x^2+3x: f(x+Δx)−f(x)=4xΔx+2(Δx)2+3Δxf(x+\Delta x)-f(x) = 4x\Delta x + 2(\Delta x)^2 + 3\Delta x.

Step 3 — Divide by Δx\Delta x: 4xΔx+2(Δx)2+3ΔxΔx=4x+2Δx+3\dfrac{4x\Delta x + 2(\Delta x)^2+3\Delta x}{\Delta x} = 4x + 2\Delta x + 3.

Step 4 — Let Δx→0\Delta x \to 0: f′(x)=4x+3f'(x) = 4x+3.

Check (independent method — power/sum rule): ddx(2x2)=4x\dfrac{d}{dx}(2x^2) = 4x and ddx(3x)=3\dfrac{d}{dx}(3x)=3, so f′(x)=4x+3f'(x)=4x+3 — matches the first-principles result exactly.

✓Final answer

f′(x)=4x+3f'(x) = 4x+3

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