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How would you differentiate ln(x^2+3x+5)?

Here we need to use the chain rule because we have a function (natural log) of another function (x^2+3x+5). Let u=x^2+3x+5, and differentiate lnu with respect to u, this gives us 1/u. Then we differentiate x^2+3x+5 with respect to x, so we get 2x+3. Now the chain rule says: dy/dx=dy/du*du/dx, so we have dy/dx = (1/u)*(2x+3)=(2x+3)/(x^2+3x+5)

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1 year ago

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