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How do you find the derivative of #y=log_4 2x#?

Answer 1

Charles Anctil

#y'=1/(xln4)#

Rewrite #y# using the change of base formula.

#y=ln(2x)/ln4#

#y'=1/ln4*d/dx(ln(2x))#

Recall that #1/ln4# is just a constant.

To find #d/dx=(ln(2x))#, use the chain rule.

#ln(u)=1/u*(du)/dx#

Thus,

#d/dx(ln(2x))=1/(2x)d/dx(2x)=2/(2x)=1/x#

Plug this back into the #y'# expression.

#y'=1/(xln4)#

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Answer 2

Luke Alvarez

To find the derivative of ( y = \log_4(2x) ), you can use the chain rule. The derivative of ( \log_4(2x) ) with respect to ( x ) is ( \frac{1}{(2x)\ln(4)} \cdot 2 ). Simplifying, you get ( \frac{1}{x\ln(4)} ). Therefore, the derivative of ( y = \log_4(2x) ) with respect to ( x ) is ( \frac{1}{x\ln(4)} ).

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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.