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How do you find the derivative of #x^(5/3) * ln(3x)#?

Answer 1

#x^(2/3)(1+5/3ln(3x))#

differentiate using the #color(blue)"product rule"#

If f(x) = g(x).h(x) then f'(x) = g(x)h'(x) + h(x)g'(x)...(A) #"-----------------------------------------------------"#

#g(x)=x^(5/3)rArrg'(x)=5/3x^(2/3)-"using the"color(blue)" power rule"# #h(x)=ln(3x)rArrh'(x)=1/(3x)xx3=1/x#

using #d/dx(ln(f(x)))=1/(f(x))xxd/dxf(x)# #"---------------------------------------------------------"# Substitute these values into (A)

#f'(x)=x^(5/3)xx1/x+ln(3x)xx5/3x^(2/3)#

#=x^(5/3-1)+5/3x^(2/3)ln(3x)=x^(2/3)+5/3x^(2/3)ln(3x)#

Taking out a common factor of #x^(2/3)#

#rArrd/dx(x^(5/3).ln(3x))=x^(2/3)(1+5/3ln(3x))#

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Answer 2

Emily Ammerman

To find the derivative of (x^\frac{5}{3} \cdot \ln(3x)), you can use the product rule and the chain rule of differentiation. The derivative is:

[ \frac{d}{dx}\left(x^\frac{5}{3} \cdot \ln(3x)\right) = \frac{5}{3}x^\frac{2}{3} \cdot \ln(3x) + x^\frac{5}{3} \cdot \frac{1}{x} ]

Simplified:

[ \frac{5}{3}x^\frac{2}{3} \cdot \ln(3x) + x^\frac{2}{3} ]

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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.