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How do you use the chain rule to differentiate #ln(tanx)#?

Answer 1

Isaiah Allen

#(dy)/(dx)=(sec^2x)/tanx#

We can treat #f(x)=ln(tanx)# as a function in terms of #tanx#.

Let #u=tanx#, #y=lnu#

Chain rule: #(dy)/(dx)=(dy)/(du)*(du)/(dx)#

#(dy)/(du)=1/u#

#(du)/(dx)=sec^2x#

#:.(dy)/(dx)=1/(tanx)*sec^2x#

#=(sec^2x)/(tanx)#

Further simplifying...

#=(1/(cos^2x))/((sinx)/(cosx))#

#=1/(sinxcosx)#

#=2/(2sinxcosx)#

#=2/(sin2x)#

#=2csc2x#

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Answer 2

Ezekiel Alexander

To differentiate ln(tanx) using the chain rule, first, you recognize that ln(u) is the natural logarithm of the function u. Then, apply the chain rule, which states that the derivative of ln(u) with respect to x is (1/u) * (du/dx). In this case, u = tanx. The derivative of tanx with respect to x is sec^2(x). So, the derivative of ln(tanx) with respect to x is (1/tanx) * sec^2(x). Simplifying, this becomes sec^2(x)/tanx.

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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.