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How do you find the derivative of #ln (1 - x^2)#?

Answer 1

Evelyn Agard

Derivative of #ln(1-x^2)# is #-(2x)/(1-x^2)#

We use te concept of function of a function. If we have #f(g(x))#, then

#(df)/(dx)=(df)/(dg(x))xx(dg)/(dx)#

Here we can write #f(x)=ln(1-x^2)# as #f(x)=ln(g(x))#, where #g(x)=1-x^2#

Now as #f(x)=ln(g(x))#, #(df)/(dg(x))=1/(g(x))#

and #(dg)/(dx)=-2x# and hence

#(df)/(dx)=1/(g(x))xx(-2x)=(-2x)/(g(x))=-(2x)/(1-x^2)#

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Answer 2

Elijah Abram

To find the derivative of ln(1 - x^2), you would use the chain rule. The derivative is:

d/dx [ln(1 - x^2)] = -2x / (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.