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What is the derivative of #f(x)=ln(secx)#?

Answer 1

#tan(x)#

We can use the chain rule here and substitute the inside of the ln function as u. So:

#ln(u), u=sec(x)#

We know the derivative of a ln function is in the form of #(u')/u#. So we need to find the derivative of #sec(x)#. This is a trig identity and so #(d(u))/dx=(d(sec(x)))/dx=sec(x)tan(x)=u' #

Putting this into our equation for the derivative of an ln function we get:

#(u')/u=(sec(x)tan(x))/sec(x)=tan(x)#

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Answer 2

Isaiah Allen

To find the derivative of ( f(x) = \ln(\sec(x)) ), you would apply the chain rule:

[ f'(x) = \frac{d}{dx}[\ln(\sec(x))] = \frac{1}{\sec(x)} \cdot \frac{d}{dx}[\sec(x)] ]

Now, differentiate ( \sec(x) ) with respect to ( x ):

[ \frac{d}{dx}[\sec(x)] = \sec(x) \tan(x) ]

Substitute this result back into the original expression:

[ f'(x) = \frac{1}{\sec(x)} \cdot \sec(x) \tan(x) = \tan(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.