differentiate wrt x, x^ln(2x^2 +3)
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differentiate x^ln(2x^2+3)  wrt x.

Let u = ln(2x^2 + 3), and y = x^u

For y = x^u, take logs of both sides, giving

ln(y) = u.ln(x)

now differentiate both sides wrt x.

(1/y)y' = u'.ln(x) + u/x   and u' = du/dx = 4x/(2x^2+3)

y' = y{[4x/(2x^2+3)].ln(x) + ln(2x^2+3)/x}

y' = x^ln(2x^2+3){4x.ln(x)/(2x^2+3) + ln(2x^2+3)/x}

by Level 11 User (80.5k points)

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