How do you differentiate # y= sqrt((3x)/(2x-3))# using the chain rule?

Answer 1

The final answer, using the chain rule and the quotient rule is:

#f'(x)=9*(2x-3)^(5/2)/(2*(3x)^(1/2)#.

See below for the details.

I find this type of function much easier to handle without the radical sign, so I will rewrite it as #f(x)=((3x)/(2x-3))^(1/2)#.

As usual with the chain rule, we begin with the "outermost" function (i.e., ask yourself, "If I wanted to find f(2), for example, what is the last function that I would do?). In this case, that is clearly the rational exponent (formerly the radical). Thus, I differentiate that first using the power rule, then handle the rational function using the quotient rule.

So: the answer becomes:

#f'(x)=(1/2)*((3x)/(2x-3))^(-1/2)*((2x-3)*3-(3x*2))/(2x-3)^2#

#f'(x)=(1/2)*((3x)/(2x-3))^(-1/2)*(6x-9-6x)/(2x-3)^2#

#f'(x)=(1/2)*(3x)^(-1/2)/(2x-3)^(-1/2)*9/(2x-3)^2#

#f'(x)=9*(2x-3)^(5/2)/(2*(3x)^(1/2)#

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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.