Question #6b55e

2 Answers
Aug 23, 2017

I will assume that we want to find #dy/dx#

Explanation:

#y = ln(tan^2x)#

Use #d/dx(lnu) = 1/u (du)/dx# (chain rule)

And the chain rule again

#d/dx(tan^2u) = d/dx((tanx)^2) = 2tanx d/dx(tanx) = 2tanxsec^2x#

So,

#dy/dx = 1/(tan^2x) d/dx(tan^2x)#

# = cot^2x (2tanxsec^2x)#

# = 2cotxsec^2x#

# = 2cscxsecx#

Aug 24, 2017

#dy/dx=2cscxsecx#

Explanation:

We can also simplify the function using the following logarithm rules:

  • #log(a^b)=blog(a)#
  • #log(a/b)=log(a)-log(b)#

So:

#y=ln(tan^2x)=2ln(tanx)=2ln(sinx/cosx)=2ln(sinx)-2ln(cosx)#

Taking the derivative, we need to remember that #d/dxlnx=1/x#. We also need to use the chain rule:

#dy/dx=2/sinx(d/dxsinx)-2/cosx(d/dxcosx)#

#dy/dx=(2cosx)/sinx+(2sinx)/cosx=(2(cos^2x+sin^2x))/(sinxcosx)#

Since #sin^2x+cos^2x=1#:

#dy/dx=2cscxsecx#