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How do you find the derivative of # f (x) = ln(1 - sin x)#?

Answer 1

Scarlett Allison

#f'(x)=cosx/(sinx-1)#

Differentiate using the #color(blue)" chain rule " #

# d/dx [ f(g(x)) ] = f'(g(x)) . g'(x) #

and # d/dx(lnx) = 1/x #

# rArr f'(x) = 1/(1 - sinx) . d/dx( 1 - sinx) = - cosx/(1 - sinx)=cosx/(sinx-1)#

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Answer 2

Charles Anctil

To find the derivative of ( f(x) = \ln(1 - \sin(x)) ), you can use the chain rule. The derivative of ( \ln(u) ) with respect to ( u ) is ( \frac{1}{u} ), and the derivative of ( u ) with respect to ( x ) is ( \frac{du}{dx} ). So, applying the chain rule, the derivative of ( \ln(1 - \sin(x)) ) with respect to ( x ) is ( \frac{1}{1 - \sin(x)} ) times the derivative of ( 1 - \sin(x) ) with respect to ( x ).

The derivative of ( 1 - \sin(x) ) with respect to ( x ) is ( -\cos(x) ). So, the derivative of ( f(x) = \ln(1 - \sin(x)) ) with respect to ( x ) is ( \frac{-\cos(x)}{1 - \sin(x)} ).

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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.