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antiseptic1488 [7]
3 years ago
10

Differentiate x^2 e^x logx

Mathematics
1 answer:
zimovet [89]3 years ago
8 0

Product rule:

\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)

=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}

Power rule:

\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x

The exponential function is its own derivative:

\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x

Assuming the base of \log x is e, its derivative is

\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x

But if you mean a logarithm of arbitrary base b, we have

y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}

\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}

So we end up with

2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x

=xe^x(2\log x+x\log x+1)

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