How do you find the derivative of #y=ln(sqrt(x))#?

Answer 1

Paisley Johnson

#frac{dy}{dx}=frac{1}{2x}#

#y=ln(sqrt(x))=ln(x^{1/2})=1/2ln(x)#

#frac{dy}{dx}=frac{1}{2x}#

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Answer 2

Elijah Abram

#y'=1/(2x)#

Explanation

#y=ln(sqrtx)#

differentiating with respect to #x# using Chain Rule,

#y'=1/sqrt(x)*1/2x^(-1/2)#

#y'=1/sqrt(x)*(1/(2sqrt(x)))#

#y'=1/(2x)#

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Answer 3

Paisley Johnson

To find the derivative of ( y = \ln(\sqrt{x}) ), you can use the chain rule.

First, rewrite the function using the power rule for logarithms: ( y = \frac{1}{2}\ln(x) ).
Now differentiate ( y ) with respect to ( x ), applying the chain rule and the derivative of the natural logarithm function: [ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x} \cdot \frac{dx}{dx} ]
Since ( \frac{dx}{dx} = 1 ), the expression simplifies to: [ \frac{dy}{dx} = \frac{1}{2x} ]

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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.