Question

differentiate by chain rule :- y = log e ( x² - 1 ) ?​

Answer 1

Answer:

The method of differentiating functions by first taking logarithms and then differentiating is called logarithmic differentiation. We use logarithmic differentiation in situations where it is easier to differentiate the logarithm of a function than to differentiate the function itself. This approach allows calculating derivatives of power, rational and some irrational functions in an efficient manner.

The method of differentiating functions by first taking logarithms and then differentiating is called logarithmic differentiation. We use logarithmic differentiation in situations where it is easier to differentiate the logarithm of a function than to differentiate the function itself. This approach allows calculating derivatives of power, rational and some irrational functions in an efficient manner.Consider this method in more detail. Let y = f\left( x \right). Take natural logarithms of both sides:

The method of differentiating functions by first taking logarithms and then differentiating is called logarithmic differentiation. We use logarithmic differentiation in situations where it is easier to differentiate the logarithm of a function than to differentiate the function itself. This approach allows calculating derivatives of power, rational and some irrational functions in an efficient manner.Consider this method in more detail. Let y = f\left( x \right). Take natural logarithms of both sides:\ln y = \ln f\left( x \right).

Answer 2

Answer:

Question :- y = log e ( x² - 1 ) ?

To find :- derivative of y = log e ( x²-1 ) .

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FORMULA USED :- y = log e^u .

☆ here, 'u' is the function of 'X'

☆ dy/dx = d/dx ( log e^u )

☆ dy/dx = 1/u ( du/dx ) .

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Answer :- y = log e (x²-1)

==> dy / dx = d/dx log e ( x² - 1 )

==> dy/dx = (1 / x²-1 ) d/dx ( x² - 1 )

==> dy/dx = ( 1/x²-1 ) . ( 2x - 0 )

==> dy/dx = 2x / x² - 1 .