Question #e8ba4

1 Answer
Andrea S.
Jan 24, 2017

#d/(dx) (sinx)^tanx = (sinx)^tanx (ln (sinx)/cos^2x +1)#

Explanation:

In cases like this the easiest approach is to pass through logarithms:

#(sinx)^tanx = (e^ln (sinx) )^tanx = e^(tanx*ln(sinx))#

so:

#d/(dx) (sinx)^tanx = e^(tanx*ln(sinx))* d/dx(tanx*ln(sinx))#

#d/(dx) (sinx)^tanx = (sinx)^tanx (ln (sinx)/cos^2x + tanx cosx/sinx)#

#d/(dx) (sinx)^tanx = (sinx)^tanx (ln (sinx)/cos^2x +1)#

Related questions
See all questions in Differentiating Exponential Functions with Other Bases
Impact of this question
1424 views around the world
You can reuse this answer
Creative Commons License