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EXERCISE 9.1 · Q3

Q.e2x+1e^{2x+1}

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✓ Free question

Let f(x)=e2x+1f(x)=e^{2x+1}, so f(x+h)=e2x+2h+1=e2x+1⋅e2hf(x+h)=e^{2x+2h+1}=e^{2x+1}\cdot e^{2h}.

f(x+h)−f(x)=e2x+1(e2h−1)f(x+h)-f(x)=e^{2x+1}\left(e^{2h}-1\right).

f(x+h)−f(x)h=e2x+1⋅e2h−1h=e2x+1⋅2⋅e2h−12h\dfrac{f(x+h)-f(x)}{h}=e^{2x+1}\cdot\dfrac{e^{2h}-1}{h}=e^{2x+1}\cdot2\cdot\dfrac{e^{2h}-1}{2h}.

As h→0h\to0: →e2x+1⋅2⋅1\to e^{2x+1}\cdot2\cdot1.

✓Final answer

f′(x)=2e2x+1f'(x)=2e^{2x+1}

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