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

Answer 1

Christopher Agresta

#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!

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Answer 2

Emily Ammerman

To differentiate the function ( y = x^2 \ln(2x) + x \ln(3x) + 4 \ln x ), apply the product rule and the chain rule:

( \frac{dy}{dx} = \frac{d}{dx}[x^2 \ln(2x)] + \frac{d}{dx}[x \ln(3x)] + \frac{d}{dx}[4 \ln x] )

Using the product rule and the chain rule, differentiate each term separately:

( \frac{dy}{dx} = (2x \ln(2x) + x^2 \cdot \frac{1}{x} \cdot 2) + (\ln(3x) + x \cdot \frac{1}{3x} \cdot 3) + \frac{4}{x} )

Simplify each term:

( \frac{dy}{dx} = 2x \ln(2x) + 2x + \ln(3x) + 1 + \frac{4}{x} )

This is the derivative of the given function.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.