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How do I find the derivative of #ln sqrt(x^2-4)#?

Answer 1

Aurora Alvarado

We'll need the chain rule, which states that #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#

Renaming #u=sqrt(v)# and #v=x^2-4#, let's do it step-by-step:

#(dy)/(du)=1/u#

#(du)/(dv)=1/(2sqrt(v))#

#(dv)/(dx)=2x#

Aggregating:

#(dy)/(dx)=(2x)/(2usqrt(v))#

Substituting #u# and then #v# afterwards:

#(dy)/(dx)=(2x)/(2(sqrt(x^2-4))(sqrt(x^2-4)))=(cancel(2)x)/(cancel(2)(x^2-4))#

#(dy)/(dx)=x/(x^2-4)#

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Answer 2

Charles Arocha

To find the derivative of ln(sqrt(x^2 - 4)), you can use the chain rule.

Let u = sqrt(x^2 - 4). Then, apply the chain rule:

Find the derivative of u with respect to x.
Find the derivative of ln(u) with respect to u.
Multiply the derivatives found in steps 1 and 2.

Derivative of u with respect to x: du/dx = (1/2)(x^2 - 4)^(-1/2) * 2x = x / sqrt(x^2 - 4)

Derivative of ln(u) with respect to u: d/dx ln(u) = 1/u

Therefore, the derivative of ln(sqrt(x^2 - 4)) with respect to x is:

d/dx ln(sqrt(x^2 - 4)) = (1 / sqrt(x^2 - 4)) * (x / sqrt(x^2 - 4)) = x / (x^2 - 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.