How do you differentiate #f(x)=ln(sinx)/cosx# using the quotient rule?

Answer 1

#f'(x)=(cosxcotx+sinxln(sinx))/cos^2x#

The quotient rule states that

#d/dx[g(x)/(h(x))]=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2#
When this is applied to #f(x)#, the function at hand, we see that
#f'(x)=(cosxd/dx[ln(sinx)]-ln(sinx)d/dx[cosx])/cos^2x#

Find each derivative separately:

To differentiate the natural logarithm function, use the chain rule, which for a natural logarithm function states that

#d/dx[ln(k(x))]=1/(k(x))*k'(x)#
Here, #k(x)=sinx#, so
#d/dx[ln(sinx)]=1/sinx*d/dx[sinx]=1/sinx*cosx=cotx#

As for the other derivative,

#d/dx[cosx]=-sinx#

Plugging these both back in, we see that

#f'(x)=(cosx(cotx)-ln(sinx)(-sinx))/cos^2x#

This can be written as

#f'(x)=(cosxcotx+sinxln(sinx))/cos^2x#

As with many trigonometric functions, this can be rewritten in many ways, including

#f'(x)=secx(cotx+tanxln(sinx))#
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Answer 2

To differentiate ( f(x) = \frac{\ln(\sin x)}{\cos x} ) using the quotient rule:

  1. Apply the quotient rule, which states that for a function ( \frac{u}{v} ), the derivative is given by ( \frac{u'v - uv'}{v^2} ).
  2. Let ( u = \ln(\sin x) ) and ( v = \cos x ).
  3. Compute ( u' ) and ( v' ) as the derivatives of ( u ) and ( v ) with respect to ( x ), respectively.
  4. Substitute ( u' ), ( v' ), ( u ), and ( v ) into the quotient rule formula.
  5. Simplify the expression.

Therefore, the derivative ( f'(x) ) of ( f(x) ) with respect to ( x ) is given by:

[ f'(x) = \frac{\frac{d}{dx}(\ln(\sin x))\cdot \cos x - \ln(\sin x) \cdot \frac{d}{dx}(\cos x)}{\cos^2 x} ]

Then, compute the derivatives of ( \ln(\sin x) ) and ( \cos x ) with respect to ( x ) using the chain rule and simplify the expression.

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Answer 3

To differentiate ( f(x) = \frac{\ln(\sin(x))}{\cos(x)} ) using the quotient rule:

  1. Identify ( u ) and ( v ) as the numerator and denominator, respectively. ( u = \ln(\sin(x)) ) and ( v = \cos(x) ).

  2. Apply the quotient rule: [ f'(x) = \frac{u'v - uv'}{v^2} ]

  3. Find the derivatives of ( u ) and ( v ): [ u' = \frac{1}{\sin(x)} \cdot \cos(x) = \frac{\cos(x)}{\sin(x)} ] [ v' = -\sin(x) ]

  4. Substitute ( u ), ( v ), ( u' ), and ( v' ) into the quotient rule formula: [ f'(x) = \frac{\frac{\cos(x)}{\sin(x)}\cos(x) - \ln(\sin(x))(-\sin(x))}{\cos^2(x)} ]

  5. Simplify the expression: [ f'(x) = \frac{\cos^2(x) + \ln(\sin(x))\sin(x)}{\sin(x)\cos^2(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.

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.

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.

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.

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