How do you differentiate $y=x^2ln(2x)+xln(3x)+4lnx$ ?

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Answer 1 · 2017-05-04T15:46:51

$y' = (2x^2ln(2x) + x^2 + xln(3x) + x + 4)/x$

Use the product and chain rules.

Developing a formula for the derivative of $ln(ax)$ , where $a$ is a constant

By the chain rule, letting $y = ln u$ and $u =ax$ . Then $du = a dx$ and $dy = 1/u du$ . We have:

$dy/dx= 1/u * a$

$dy/dx = a/(ax)$

$dy/dx = 1/x$

Apply the formula to the problem

$y' = 2xln(2x) + x^2(1/x) + ln(3x) + x(1/x) + 4/x$

$y' = 2xln(2x) + x + ln(3x) + 1 + 4/x$

$y' = (2x^2ln(2x) + x^2 + xln(3x) + x + 4)/x$

Hopefully this helps!

How do you differentiate y=x^2ln(2x)+xln(3x)+4lnxy=x^2ln(2x)+xln(3x)+4lnx?