How do you find the first and second derivative of #ln(x/2)#?

Answer 1

Samantha Adkison

# dy/dx= 1/x#
# (d^2y)/(dx^2) = -1/x^2 #

Let # y = ln(x/2) #

We can use the law of logarithms to write:

# y= lnx - ln2 #

Differentiating wrt #x# we have;

# dy/dx= 1/x#

We can rewrite this as:

# dy/dx= x^-1 #

Differentiating again wrt #x# we have;

# (d^2y)/(dx^2) = (-1)x^-2 #

# (d^2y)/(dx^2) = -1/x^2 #

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Answer 2

Ezra Adams

To find the first derivative of ( \ln(x/2) ), you use the chain rule: [ \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} ]

Now, let ( u = x/2 ): [ \frac{du}{dx} = \frac{d}{dx} \left(\frac{x}{2}\right) = \frac{1}{2} ]

Substituting back into the chain rule formula: [ \frac{d}{dx} \ln(x/2) = \frac{1}{x/2} \cdot \frac{1}{2} = \frac{1}{x} ]

To find the second derivative, differentiate ( \frac{1}{x} ) with respect to ( x ): [ \frac{d^2}{dx^2} \left( \frac{1}{x} \right) = -\frac{1}{x^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.