Question #13d2c

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Answer 1 · 2016-10-04T19:42:47

You'll need some form of logarithmic differentiation.

The work is the same, but there are two ways to think about this.

Method 1
#y = (secx)^x#, so

#lny = ln((secx)^x) = xln(secx)#.

Differentiate implicitly:

#1/y dy/dx = 1 ln(secx) + x[1/secx secx tanx]#
#dy/dx = (secx)^x(lnsecx + x tanx)#

Method 2

#y = e^ln((secx)^x) = e^(xln(secx))#

#dy/dx = e^(xln(secx))[d/dx(xln(secx))]#

# = e^(xln(secx))[ 1 ln(secx) + x(1/secx secx tanx) ]#

# = (secx)^x[ln(secx) +x tanx]#